cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where m
Between every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

Examples

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
  • L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E22.
  • A. M. Hinz, The Tower of Hanoi, in Algebras and combinatorics (Hong Kong, 1997), 277-289, Springer, Singapore, 1999.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589.
  • Andrew Schloss, "Towers of Hanoi" composition, in The Digital Domain. Elektra/Asylum Records 9 60303-2, 1983. Works by Jaffe (Finale to "Silicon Valley Breakdown"), McNabb ("Love in the Asylum"), Schloss ("Towers of Hanoi"), Mattox ("Shaman"), Rush, Moorer ("Lions are Growing") and others.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A000005, A000041, A000079, A003188, A003278, A003602, A007949, A018238, A047999, A051731, A054525, A054852, A065176, A082850, A089080, A092119, A117303, A129360, A130093, A173238, A181988, A220466.
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

Programs

  • Haskell
    a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
-- Reinhard Zumkeller, Sep 27 2011
  • Haskell
    a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
-- Walt Rorie-Baety, Mar 22 2013
  • MATLAB
    nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016
  • Magma
    [Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015
  • Maple
    A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
# This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018
  • Mathematica
    Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)
  • PARI
    a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */
  • PARI
    a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */
  • PARI
    {a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */
  • PARI
    {a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */
  • Python
    def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017
  • Python
    A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020
  • Python
    def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022
  • Sage
    [valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015
  • Scheme
    (define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Extensions

Name edited following suggestion by David James Sycamore, Oct 05 2023

A003056 n appears n+1 times. Also the array A(n,k) = n+k (n >= 0, k >= 0) read by antidiagonals. Also inverse of triangular numbers.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also triangle read by rows: T(n,k), n>=0, k>=0, in which n appears n+1 times in row n. - Omar E. Pol, Jul 15 2012
The PARI functions t1, t2 can be used to read a triangular array T(n,k) (n >= 0, 0 <= k <= n-1) by rows from left to right: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Number of terms in partition of n with greatest number of distinct terms. - Amarnath Murthy, May 20 2001
Summation table for (x+y) = (0+0),(0+1),(1+0),(0+2),(1+1),(2+0), ...
Also the number of triangular numbers less than or equal to n, not counting 0 as triangular. - Robert G. Wilson v, Oct 21 2005
Permutation of A116939: a(n) = A116939(A116941(n)), a(A116942(n)) = A116939(n). - Reinhard Zumkeller, Feb 27 2006
Maximal size of partitions of n into distinct parts, see A000009. - Reinhard Zumkeller, Jun 13 2009
Also number of digits of A000462(n). - Reinhard Zumkeller, Mar 27 2011
Also the maximum number of 1's contained in the list of hook-lengths of a partition of n. E.g., a(4)=2 because hooks of partitions of n=4 comprise {4,3,2,1}, {4,2,1,1}, {3,2,2,1}, {4,1,2,1}, {4,3,2,1} where the number of 1's in each is 1,2,1,2,1. Hence the maximum is 2. - T. Amdeberhan, Jun 03 2012
Fan, Yang, and Yu (2012) prove a conjecture of Amdeberhan on the generating function of a(n). - Jonathan Sondow, Dec 17 2012
Also the number of partitions of n into distinct parts p such that max(p) - min(p) <= length(p). - Clark Kimberling, Apr 18 2014
Also the maximum number of occurrences of any single value among the previous terms. - Ivan Neretin, Sep 20 2015
Where records occur gives A000217. - Omar E. Pol, Nov 05 2015
Also number of peaks in the largest Dyck path of the symmetric representation of sigma(n), n >= 1. Cf. A237593. - Omar E. Pol, Dec 19 2016

Examples

			G.f. = x + x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + ...
As triangle, the sequence starts
  0;
  1, 1;
  2, 2, 2;
  3, 3, 3, 3;
  4, 4, 4, 4, 4;
  5, 5, 5, 5, 5, 5;
  6, 6, 6, 6, 6, 6, 6;
  7, 7, 7, 7, 7, 7, 7, 7;
  8, 8, 8, 8, 8, 8, 8, 8, 8;
  ...

Links

Crossrefs

a(n) = A002024(n+1)-1.
Cf. A000196, A000217, A000462, A001227, A001462, A001614, A004247 (multiplication table), A006463 (partial sums), A016655, A050600, A050602, A048645, A122797, A131507, A238005.
Partial sums of A073424.
Cf. A116939, A116941, A116942,

Programs

  • Haskell
    a003056 = floor . (/ 2) . (subtract 1) .
                  sqrt . (+ 1) . (* 8) . fromIntegral
a003056_row n = replicate (n + 1) n
a003056_tabl = map a003056_row [0..]
a003056_list = concat $ a003056_tabl
-- Reinhard Zumkeller, Aug 02 2014, Oct 17 2010
  • Magma
    [Floor((Sqrt(1+8*n)-1)/2): n in [0..80]]; // Vincenzo Librandi, Oct 23 2011
  • Maple
    A003056 := (n,k) -> n: # Peter Luschny, Oct 29 2011
a := [ 0 ]: for i from 1 to 15 do for j from 1 to i+1 do a := [ op(a),i ]; od: od: a;
A003056 := proc(n)
    floor((sqrt(1+8*n)-1)/2) ;
end proc: # R. J. Mathar, Jul 10 2015
  • Mathematica
    f[n_] := Floor[(Sqrt[1 + 8n] - 1)/2]; Table[ f[n], {n, 0, 87}] (* Robert G. Wilson v, Oct 21 2005 *)
Table[x, {x, 0, 13}, {y, 0, x}] // Flatten
T[ n_, k_] := If[ n >= k >= 0, n, 0]; (* Michael Somos, Dec 22 2016 *)
Flatten[Table[PadRight[{},n+1,n],{n,0,12}]] (* Harvey P. Dale, Jul 03 2021 *)
  • PARI
    A003056(n)=(sqrtint(8*n+1)-1)\2  \\ M. F. Hasler, Oct 08 2011
  • PARI
    t1(n)=floor(-1/2+sqrt(2+2*n)) /* A003056 */
  • PARI
    t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */
  • Python
    from math import isqrt
def A003056(n): return (k:=isqrt(m:=n+1<<1))+int((m<<2)>(k<<2)*(k+1)+1)-1 # Chai Wah Wu, Jul 26 2022

Formula

a(n) = floor((sqrt(1+8*n)-1)/2). - Antti Karttunen
a(n) = floor(-1/2 + sqrt(2*n+b)) with 1/4 <= b < 9/4 or a(n) = floor((sqrt(8*n+b)-1)/2) with 1 <= b < 9. - Michael A. Childers (childers_moof(AT)yahoo.com), Nov 11 2001
a(n) = f(n,0) with f(n,k) = k if n <= k, otherwise f(n-k-1, k+1). - Reinhard Zumkeller, May 23 2009
a(n) = 2*n + 1 - A001614(n+1) = n + 1 - A122797(n+1). - Reinhard Zumkeller, Feb 12 2012
a(n) = k if k*(k+1)/2 <= n < (k+1)*(k+2)/2. - Jonathan Sondow, Dec 17 2012
G.f.: (1-x)^(-1)*Sum_{n>=1} x^(n*(n+1)/2) = (Theta_2(0,x^(1/2)) - 2*x^(1/8))/(2*x^(1/8)*(1-x)) where Theta_2 is a Jacobi Theta function. - Robert Israel, May 21 2015
a(n) = floor((A000196(1+8*n)-1)/2). - Pontus von Brömssen, Dec 10 2018
a(n+1) = a(n-a(n)) + 1, a(0) = 0. - Rok Cestnik, Dec 29 2020
a(n) = A001227(n) + A238005(n), n >= 1. - Omar E. Pol, Sep 30 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 (cf. A016655). - Amiram Eldar, Sep 24 2023
G.f. as array: (x + y - 2*x*y)/((1 - x)^2*(1 - y)^2). - Stefano Spezia, Dec 20 2023 [corrected by Stefano Spezia, Apr 22 2024]

Extensions

Definition clarified by N. J. A. Sloane, Dec 08 2020

A004736 Triangle read by rows: row n lists the first n positive integers in decreasing order.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 1, 6, 5, 4, 3, 2, 1, 7, 6, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
Offset: 1

Views

Author

R. Muller

Keywords

Comments

Old name: Triangle T(n,k) = n-k, n >= 1, 0 <= k < n. Fractal sequence formed by repeatedly appending strings m, m-1, ..., 2, 1.
The PARI functions t1 (this sequence), t2 (A002260) can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002, edited by M. F. Hasler, Mar 31 2020
A004736 is the mirror of the self-fission of the polynomial sequence (q(n,x)) given by q(n,x) = x^n+ x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Seen as flattened list: a(A000217(n)) = 1; a(A000124(n)) = n and a(m) <> n for m < A000124(n). - Reinhard Zumkeller, Jul 22 2012
Sequence B is called a reverse reluctant sequence of sequence A, if B is triangle array read by rows: row number k lists first k elements of the sequence A in reverse order. Sequence A004736 is the reverse reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 13 2012
The row sums equal A000217(n). The alternating row sums equal A004526(n+1). The antidiagonal sums equal A002620(n+1) respectively A008805(n-1). - Johannes W. Meijer, Sep 28 2013
From Peter Bala, Jul 29 2014: (Start)
Riordan array (1/(1-x)^2,x). Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the infinite matrix product M(0)*M(1)*M(2)*... is equal to A078812. (End)
T(n, k) gives the number of subsets of [n] := {1, 2, ..., n} with k consecutive numbers (consecutive k-subsets of [n]). - Wolfdieter Lang, May 30 2018
a(n) gives the distance from (n-1) to the smallest triangular number > (n-1). - Ctibor O. Zizka, Apr 09 2020
To construct the sequence, start from 1,2,,3,,,4,,,,5,,,,,6... where there are n commas after each "n". Then fill the empty places by the sequence itself. - Benoit Cloitre, Aug 17 2021
T(n,k) is the number of cycles of length 2*(k+1) in the (n+1)-ladder graph. There are no cycles of odd length. - Mohammed Yaseen, Jan 14 2023
The first 77 entries are also the signature sequence of log(3)=A002391. Then the two sequences start to differ. - R. J. Mathar, May 27 2024

Examples

			The triangle T(n, k) starts:
   n\k  1  2  3  4  5  6  7  8  9 10 11 12 ...
   1:   1
   2:   2  1
   3:   3  2  1
   4:   4  3  2  1
   5:   5  4  3  2  1
   6:   6  5  4  3  2  1
   7:   7  6  5  4  3  2  1
   8:   8  7  6  5  4  3  2  1
   9:   9  8  7  6  5  4  3  2  1
  10:  10  9  8  7  6  5  4  3  2  1
  11:  11 10  9  8  7  6  5  4  3  2  1
  12:  12 11 10  9  8  7  6  5  4  3  2  1
  ... Reformatted. - _Wolfdieter Lang_, Feb 04 2015
T(6, 3) = 4 because the four consecutive 3-subsets of [6] = {1, 2, ..., 6} are {1, 2, 3}, {2, 3, 4}, {3, 4, 5} and {4, 5, 6}. - _Wolfdieter Lang_, May 30 2018

References

  • H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp 159-162.

Links

Crossrefs

Cf. A000217, A002024, A002262, A003056, A025581.
Ordinal transform of A002260. See also A078812.
Cf. A141419 (partial sums per row).
Cf. A134546 (T * A051731, matrix product).
See A001511 for definition of ordinal transform.
Cf. A128174 (parity).

Programs

  • Excel
    =if(row()>=column();row()-column()+1;"") [Mats Granvik, Jan 19 2009]
  • Haskell
    a004736 n k = n - k + 1
a004736_row n = a004736_tabl !! (n-1)
a004736_tabl = map reverse a002260_tabl
-- Reinhard Zumkeller, Aug 04 2014, Jul 22 2012
  • Maple
    A004736 := proc(n,m) n-m+1 ; end:
T := (n, k) -> n-k+1: seq(seq(T(n,k), k=1..n), n=1..13); # Johannes W. Meijer, Sep 28 2013
  • Mathematica
    Flatten[ Table[ Reverse[ Range[n]], {n, 12}]] (* Robert G. Wilson v, Apr 27 2004 *)
Table[Range[n,1,-1],{n,20}]//Flatten (* Harvey P. Dale, May 27 2020 *)
  • PARI
    {a(n) = 1 + binomial(1 + floor(1/2 + sqrt(2*n)), 2) - n}
  • PARI
    {t1(n) = binomial( floor(3/2 + sqrt(2*n)), 2) - n + 1} /* A004736 */
  • PARI
    {t2(n) = n - binomial( floor(1/2 + sqrt(2*n)), 2)} /* A002260 */
  • PARI
    apply( A004736(n)=1-n+(n=sqrtint(8*n)\/2)*(n+1)\2, [1..99]) \\ M. F. Hasler, Mar 31 2020
  • Python
    def agen(rows):
    for n in range(1, rows+1): yield from range(n, 0, -1)
print([an for an in agen(13)]) # Michael S. Branicky, Aug 17 2021
  • Python
    from math import comb, isqrt
def A004736(n): return comb((m:=isqrt(k:=n<<1))+(k>m*(m+1))+1,2)+1-n # Chai Wah Wu, Nov 08 2024

Formula

a(n+1) = 1 + A025581(n).
a(n) = (2 - 2*n + round(sqrt(2*n)) + round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003
G.f.: 1 / ((1-x)^2 * (1-x*y)). - Ralf Stephan, Jan 23 2005
Recursion: e(n,k) = (e(n - 1, k)*e(n, k - 1) + 1)/e(n - 1, k - 1). - Roger L. Bagula, Mar 25 2009
a(n) = (t*t+3*t+4)/2-n, where t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012
From Johannes W. Meijer, Sep 28 2013: (Start)
T(n, k) = n - k + 1, n >= 1 and 1 <= k <= n.
T(n, k) = A002260(n+k-1, n-k+1). (End)
a(n) = A000217(A002024(n)) - n + 1. - Enrique Pérez Herrero, Aug 29 2016

Extensions

New name from Omar E. Pol, Jul 15 2012

A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).

Original entry on oeis.org

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487, 6769
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
Crystal ball sequence for A_2 lattice. - Michael Somos, Jun 03 2012
Sixth spoke of hexagonal spiral (cf. A056105-A056109).
Number of ordered integer triples (a,b,c), -n <= a,b,c <= n, such that a+b+c=0. - Benoit Cloitre, Jun 14 2003
Also the number of partitions of 6n into at most 3 parts, A001399(6n). - R. K. Guy, Oct 20 2003
Also, a(n) is the number of partitions of 6(n+1) into exactly 3 distinct parts. - William J. Keith, Jul 01 2004
Number of dots in a centered hexagonal figure with n+1 dots on each side.
Values of second Bessel polynomial y_2(n) (see A001498).
First differences of cubes (A000578). - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
Final digits of Hex numbers (hex(n) mod 10) are periodic with palindromic period of length 5 {1, 7, 9, 7, 1}. Last two digits of Hex numbers (hex(n) mod 100) are periodic with palindromic period of length 100. - Alexander Adamchuk, Aug 11 2006
All divisors of a(n) are congruent to 1, modulo 6. Proof: If p is an odd prime different from 3 then 3n^2 + 3n + 1 = 0 (mod p) implies 9(2n + 1)^2 = -3 (mod p), whence p = 1 (mod 6). - Nick Hobson, Nov 13 2006
For n>=1, a(n) is the side of Outer Napoleon Triangle whose reference triangle is a right triangle with legs (3a(n))^(1/2) and 3n(a(n))^(1/2). - Tom Schicker (tschicke(AT)email.smith.edu), Apr 25 2007
Number of triples (a,b,c) where 0<=(a,b)<=n and c=n (at least once the term n). E.g., for n = 1: (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0), (1,1,1), so a(1)=7. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Equals the triangular numbers convolved with [1, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009
From Terry Stickels, Dec 07 2009: (Start)
Also the maximum number of viewable cubes from any one static point while viewing a cube stack of identical cubes of varying magnitude.
For example, viewing a 2 X 2 X 2 stack will yield 7 maximum viewable cubes.
If the stack is 3 X 3 X 3, the maximum number of viewable cubes from any one static position is 19, and so on.
The number of cubes in the stack must always be the same number for width, length, height (at true regular cubic stack) and the maximum number of visible cubes can always be found by taking any cubic number and subtracting the number of the cube that is one less.
Examples: 125 - 64 = 61, 64 - 27 = 37, 27 - 8 = 19. (End)
The sequence of digital roots of the a(n) is period 3: repeat [1,7,1]. - Ant King, Jun 17 2012
The average of the first n (n>0) centered hexagonal numbers is the n-th square. - Philippe Deléham, Feb 04 2013
A002024 is the following array A read along antidiagonals:
1, 2, 3, 4, 5, 6, ...
2, 3, 4, 5, 6, 7, ...
3, 4, 5, 6, 7, 8, ...
4, 5, 6, 7, 8, 9, ...
5, 6, 7, 8, 9, 10, ...
6, 7, 8, 9, 10, 11, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n) is the sum of the terms in the n+1 X n+1 matrices minus those in n X n matrices in an array formed by considering A158405 an array (the beginning terms in each row are 1,3,5,7,9,11,...). - J. M. Bergot, Jul 05 2013
The formula also equals the product of the three distinct combinations of two consecutive numbers: n^2, (n+1)^2, and n*(n+1). - J. M. Bergot, Mar 28 2014
The sides of any triangle ABC are divided into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_2n in side a, and also on the sides b and c cyclically. If A'B'C' is the triangle delimited by AA_n, BB_n and CC_n cevians, we have (ABC)/(A'B'C') = a(n) (see Java applet link). - Ignacio Larrosa Cañestro, Jan 02 2015
a(n) is the maximal number of parts into which (n+1) triangles can intersect one another. - Ivan N. Ianakiev, Feb 18 2015
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = ((2^m-1)(2n+1))^t mod a(n), where m any positive integer, and t = 0(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = a(n) - (((2^m-1)(2n+1))^t mod a(n)), where m any positive integer, and t = 3(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
(3n+1)^(a(n)-1) mod a(n) = (3n+2)^(a(n)-1) mod a(n) = 1. If a(n) not prime, then always strong pseudoprime. - Alzhekeyev Ascar M, Oct 07 2016
Every positive integer is the sum of 8 hex numbers (zero included), at most 3 of which are greater than 1. - Mauro Fiorentini, Jan 01 2018
Area enclosed by the segment of Archimedean spiral between n*Pi/2 and (n+1)*Pi/2 in Pi^3/48 units. - Carmine Suriano, Apr 10 2018
This sequence contains all numbers k such that 12*k - 3 is a square. - Klaus Purath, Oct 19 2021
The continued fraction expansion of sqrt(3*a(n)) is [3n+1; {1, 1, 2n, 1, 1, 6n+2}]. For n = 0, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 12 2022

Examples

			G.f. = 1 + 7*x + 19*x^2 + 37*x^3 + 61*x^4 + 91*x^5 + 127*x^6 + 169*x^7 + 217*x^8 + ...
From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms:
.
.                                 o o o o
.                   o o o        o o o o o
.         o o      o o o o      o o o o o o
.   o    o o o    o o o o o    o o o o o o o
.         o o      o o o o      o o o o o o
.                   o o o        o o o o o
.                                 o o o o
.
.   1      7          19             37
.
(End)
From _Klaus Purath_, Dec 03 2021: (Start)
(1) a(19) is not a prime number, because besides a(19) = a(9) + P(29), a(19) = a(15) + P(20) = a(2) + P(33) is also true.
(2) a(25) is prime, because except for a(25) = a(12) + P(38) there is no other equation of this pattern. (End)

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000124, A000166, A000217, A000290, A000578 (the cubes, or partial sums), A001263, A001498, A002061, A002378, A002407 (primes), A003514, A005408, A005449, A005891, A028896, A048766, A056105, A056106, A056107, A056108, A056109, A063496, A056220, A130298, A132111 (second diagonal), A158405, A215630, A239449, A243201.
Column k=3 of A080853, and column k=2 of A047969.
See also A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.
Cf. A287326(A000124(n), 1).
Cf. A008292.
Cf. A005448, A140091, A001844, A000384, A060544, A045943.
Cf. A154105.

Programs

Formula

a(n) = 3*n*(n+1) + 1, n >= 0 (see the name).
a(n) = (n+1)^3 - n^3 = a(-1-n).
G.f.: (1 + 4*x + x^2) / (1 - x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = 6*A000217(n) + 1.
a(n) = a(n-1) + 6*n = 2a(n-1) - a(n-2) + 6 = 3*a(n-1) - 3*a(n-2) + a(n-3) = A056105(n) + 5n = A056106(n) + 4*n = A056107(n) + 3*n = A056108(n) + 2*n = A056108(n) + n.
n-th partial arithmetic mean is n^2. - Amarnath Murthy, May 27 2003
a(n) = 1 + Sum_{j=0..n} (6*j). E.g., a(2)=19 because 1+ 6*0 + 6*1 + 6*2 = 19. - Xavier Acloque, Oct 06 2003
The sum of the first n hexagonal numbers is n^3. That is, Sum_{n>=1} (3*n*(n-1) + 1) = n^3. - Edward Weed (eweed(AT)gdrs.com), Oct 23 2003
a(n) = right term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 0 0 / 2 1 0 / 3 3 1]. M^n * [1 1 1] = [1 2n+1 a(n)]. E.g., a(4) = 61, right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 9 61] = [1 2n+1 a(4)]. - Gary W. Adamson, Dec 22 2004
Row sums of triangle A130298. - Gary W. Adamson, Jun 07 2007
a(n) = 3*n^2 + 3*n + 1. Proof: 1) If n occurs once, it may be in 3 positions; for the two other ones, n terms are independently possible, then we have 3*n^2 different triples. 2) If the term n occurs twice, the third one may be placed in 3 positions and have n possible values, then we have 3*n more different triples. 3) The term n may occurs 3 times in one way only that gives the formula. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Binomial transform of [1, 6, 6, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 6, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = (n-1)*A000166(n) + (n-2)*A000166(n-1) = (n-1)floor(n!*e^(-1)+1) + (n-2)*floor((n-1)!*e^(-1)+1) (with offset 0). - Gary Detlefs, Dec 06 2009
a(n) = A028896(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = integral( (sin((n+1/2)x)/sin(x/2))^3, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011
Sum_{n>=0} 1/a(n) = Pi/sqrt(3)*tanh(Pi/(2*sqrt(3))) = 1.305284153013581... - Ant King, Jun 17 2012
a(n) = A000290(n) + A000217(2n+1). - Ivan N. Ianakiev, Sep 24 2013
a(n) = A002378(n+1) + A056220(n) = A005408(n) + 2*A005449(n) = 6*A000217(n) + 1. - Ivan N. Ianakiev, Sep 26 2013
a(n) = 6*A000124(n) - 5. - Ivan N. Ianakiev, Oct 13 2013
a(n) = A239426(n+1) / A239449(n+1) = A215630(2*n+1,n+1). - Reinhard Zumkeller, Mar 19 2014
a(n) = A243201(n) / A002061(n + 1). - Mathew Englander, Jun 03 2014
a(n) = A101321(6,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (1 + 6*x + 3*x^2)*exp(x). - Ilya Gutkovskiy, Jul 28 2016
a(n) = (A001844(n) + A016754(n))/2. - Bruce J. Nicholson, Aug 06 2017
a(n) = A045943(2n+1). - Miquel Cerda, Jan 22 2018
a(n) = 3*Integral_{x=n..n+1} x^2 dx. - Carmine Suriano, Apr 10 2018
a(n) = A287326(A000124(n), 1). - Kolosov Petro, Oct 22 2018
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 10*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 2/e. (End)
G.f.: polylog(-3, x)*(1-x)/x. See the Simon Plouffe formula above, and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 08 2021
a(n) = T(n-1)^2 - 2*T(n)^2 + T(n+1)^2, n >= 1, T = triangular number A000217. - Klaus Purath, Oct 11 2021
a(n) = 1 + 2*Sum_{j=n..2n} j. - Klaus Purath, Oct 19 2021
a(n) = A069099(n+1) - A000217(n). - Klaus Purath, Nov 03 2021
From Leo Tavares, Dec 03 2021: (Start)
a(n) = A005448(n) + A140091(n);
a(n) = A001844(n) + A002378(n);
a(n) = A005891(n) + A000217(n);
a(n) = A000290(n) + A000384(n+1);
a(n) = A060544(n-1) + 3*A000217(n);
a(n) = A060544(n-1) + A045943(n).
a(2*n+1) = A154105(n).
(End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A002262 Triangle read by rows: T(n,k) = k, 0 <= k <= n, in which row n lists the first n+1 nonnegative integers.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
Offset: 0

Views

Author

Angele Hamel (amh(AT)maths.soton.ac.uk)

Keywords

Comments

The point with coordinates (x = A025581(n), y = A002262(n)) sweeps out the first quadrant by upwards antidiagonals. N. J. A. Sloane, Jul 17 2018
Old name: Integers 0 to n followed by integers 0 to n+1 etc.
a(n) = n - the largest triangular number <= n. - Amarnath Murthy, Dec 25 2001
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 0, k >= 0) by antidiagonals downwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Values x of unique solution pair (x,y) to equation T(x+y) + x = n, where T(k)=A000217(k). - Lekraj Beedassy, Aug 21 2004
a(A000217(n)) = 0; a(A000096(n)) = n. - Reinhard Zumkeller, May 20 2009
Concatenation of the set representation of ordinal numbers, where the n-th ordinal number is represented by the set of all ordinals preceding n, 0 being represented by the empty set. - Daniel Forgues, Apr 27 2011
An integer sequence is nonnegative if and only if it is a subsequence of this sequence. - Charles R Greathouse IV, Sep 21 2011
a(A195678(n)) = A000040(n) and a(m) <> A000040(n) for m < A195678(n), an example of the preceding comment. - Reinhard Zumkeller, Sep 23 2011
A sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. A002262 is reluctant sequence of 0,1,2,3,... The nonnegative integers, A001477. - Boris Putievskiy, Dec 12 2012

Examples

			From _Daniel Forgues_, Apr 27 2011: (Start)
Examples of set-theoretic representation of ordinal numbers:
  0: {}
  1: {0} = {{}}
  2: {0, 1} = {0, {0}} = {{}, {{}}}
  3: {0, 1, 2} = {{}, {0}, {0, 1}} = ... = {{}, {{}}, {{}, {{}}}} (End)
From _Omar E. Pol_, Jul 15 2012: (Start)
  0;
  0, 1;
  0, 1, 2;
  0, 1, 2, 3;
  0, 1, 2, 3, 4;
  0, 1, 2, 3, 4, 5;
  0, 1, 2, 3, 4, 5, 6;
  0, 1, 2, 3, 4, 5, 6, 7;
  0, 1, 2, 3, 4, 5, 6, 7, 8;
  0, 1, 2, 3, 4, 5, 6, 7, 8, 9;
  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
(End)

Links

Crossrefs

Cf. A002024, A002260, A004736, A025581, A025675, A025682.
Cf. A025691, A048645, A053186, A053645, A056558, A127324.
As a sequence, essentially same as A048151.
Cf. A060510 (parity).

Programs

  • Haskell
    a002262 n k = a002262_tabl !! n !! k
a002262_row n = a002262_tabl !! n
a002262_tabl = map (enumFromTo 0) [0..]
a002262_list = concat a002262_tabl
-- Reinhard Zumkeller, Aug 05 2015, Jul 13 2012, Mar 07 2011
  • Maple
    seq(seq(i,i=0..n),n=0..14); # Peter Luschny, Sep 22 2011
A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);
  • Mathematica
    m[n_]:= Floor[(-1 + Sqrt[8n - 7])/2]
b[n_]:= n - m[n] (m[n] + 1)/2
Table[m[n], {n, 1, 105}]     (* A003056 *)
Table[b[n], {n, 1, 105}]     (* A002260 *)
Table[b[n] - 1, {n, 1, 120}] (* A002262 *)
(* Clark Kimberling, Jun 14 2011 *)
Flatten[Table[k, {n, 0, 14}, {k, 0, n}]] (* Alonso del Arte, Sep 21 2011 *)
Flatten[Table[Range[0,n], {n,0,15}]] (* Harvey P. Dale, Jan 31 2015 *)
  • PARI
    a(n)=n-binomial(round(sqrt(2+2*n)),2)
  • PARI
    t1(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262, this sequence */
  • PARI
    t2(n)=binomial(floor(3/2+sqrt(2+2*n)),2)-(n+1) /* A025581, cf. comment by Somos for reading arrays by antidiagonals */
  • PARI
    concat(vector(15,n,vector(n,i,i-1)))  \\ M. F. Hasler, Sep 21 2011
  • PARI
    apply( {A002262(n)=n-binomial(sqrtint(8*n+8)\/2,2)}, [0..99]) \\ M. F. Hasler, Oct 20 2022
  • Python
    for i in range(16):
    for j in range(i):
        print(j, end=", ") # Mohammad Saleh Dinparvar, May 13 2020
  • Python
    from math import comb, isqrt
def a(n): return n - comb((1+isqrt(8+8*n))//2, 2)
print([a(n) for n in range(105)]) # Michael S. Branicky, May 07 2023

Formula

a(n) = A002260(n) - 1.
a(n) = n - (trinv(n)*(trinv(n)-1))/2; trinv := n -> floor((1+sqrt(1+8*n))/2) (cf. A002024); # gives integral inverses of triangular numbers
a(n) = n - A000217(A003056(n)) = n - A057944(n). - Lekraj Beedassy, Aug 21 2004
a(n) = A140129(A023758(n+2)). - Reinhard Zumkeller, May 14 2008
a(n) = f(n,1) with f(n,m) = if nReinhard Zumkeller, May 20 2009
a(n) = (1/2)*(t - t^2 + 2*n), where t = floor(sqrt(2*n+1) + 1/2) = round(sqrt(2*n+1)). - Ridouane Oudra, Dec 01 2019
a(n) = ceiling((-1 + sqrt(9 + 8*n))/2) * (1 - ((1/2)*ceiling((1 + sqrt(9 + 8*n))/2))) + n. - Ryan Jean, Sep 03 2022
G.f.: x*y/((1 - x)*(1 - x*y)^2). - Stefano Spezia, Feb 21 2024

Extensions

New name from Omar E. Pol, Jul 15 2012
Typo in definition fixed by Reinhard Zumkeller, Aug 05 2015

A001563 a(n) = n*n! = (n+1)! - n!.

Original entry on oeis.org

0, 1, 4, 18, 96, 600, 4320, 35280, 322560, 3265920, 36288000, 439084800, 5748019200, 80951270400, 1220496076800, 19615115520000, 334764638208000, 6046686277632000, 115242726703104000, 2311256907767808000, 48658040163532800000, 1072909785605898240000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

A similar sequence, with the initial 0 replaced by 1, namely A094258, is defined by the recurrence a(2) = 1, a(n) = a(n-1)*(n-1)^2/(n-2). - Andrey Ryshevich (ryshevich(AT)notes.idlab.net), May 21 2002
Denominators in power series expansion of E_1(x) + gamma + log(x), x > 0. - Michael Somos, Dec 11 2002
If all the permutations of any length k are arranged in lexicographic order, the n-th term in this sequence (n <= k) gives the index of the permutation that rotates the last n elements one position to the right. E.g., there are 24 permutations of 4 items. In lexicographic order they are (0,1,2,3), (0,1,3,2), (0,2,1,3), ... (3,2,0,1), (3,2,1,0). Permutation 0 is (0,1,2,3), which rotates the last 1 element, i.e., it makes no change. Permutation 1 is (0,1,3,2), which rotates the last 2 elements. Permutation 4 is (0,3,1,2), which rotates the last 3 elements. Permutation 18 is (3,0,1,2), which rotates the last 4 elements. The same numbers work for permutations of any length. - Henry H. Rich (glasss(AT)bellsouth.net), Sep 27 2003
Stirling transform of a(n+1)=[4,18,96,600,...] is A083140(n+1)=[4,22,154,...]. - Michael Somos, Mar 04 2004
From Michael Somos, Apr 27 2012: (Start)
Stirling transform of a(n)=[1,4,18,96,...] is A069321(n)=[1,5,31,233,...].
Partial sums of a(n)=[0,1,4,18,...] is A033312(n+1)=[0,1,5,23,...].
Binomial transform of A000166(n+1)=[0,1,2,9,...] is a(n)=[0,1,4,18,...].
Binomial transform of A000255(n+1)=[1,3,11,53,...] is a(n+1)=[1,4,18,96,...].
Binomial transform of a(n)=[0,1,4,18,...] is A093964(n)=[0,1,6,33,...].
Partial sums of A001564(n)=[1,3,4,14,...] is a(n+1)=[1,4,18,96,...].
(End)
Number of small descents in all permutations of [n+1]. A small descent in a permutation (x_1,x_2,...,x_n) is a position i such that x_i - x_(i+1) =1. Example: a(2)=4 because there are 4 small descents in the permutations 123, 13\2, 2\13, 231, 312, 3\2\1 of {1,2,3} (shown by \). a(n)=Sum_{k=0..n-1}k*A123513(n,k). - Emeric Deutsch, Oct 02 2006
Equivalently, in the notation of David, Kendall and Barton, p. 263, this is the total number of consecutive ascending pairs in all permutations on n+1 letters (cf. A010027). - N. J. A. Sloane, Apr 12 2014
a(n-1) is the number of permutations of n in which n is not fixed; equivalently, the number of permutations of the positive integers in which n is the largest element that is not fixed. - Franklin T. Adams-Watters, Nov 29 2006
Number of factors in a determinant when writing down all multiplication permutations. - Mats Granvik, Sep 12 2008
a(n) is also the sum of the positions of the left-to-right maxima in all permutations of [n]. Example: a(3)=18 because the positions of the left-to-right maxima in the permutations 123,132,213,231,312 and 321 of [3] are 123, 12, 13, 12, 1 and 1, respectively and 1+2+3+1+2+1+3+1+2+1+1=18. - Emeric Deutsch, Sep 21 2008
Equals eigensequence of triangle A002024 ("n appears n times"). - Gary W. Adamson, Dec 29 2008
Preface the series with another 1: (1, 1, 4, 18, ...); then the next term = dot product of the latter with "n occurs n times". Example: 96 = (1, 1, 4, 8) dot (4, 4, 4, 4) = (4 + 4 + 16 + 72). - Gary W. Adamson, Apr 17 2009
Row lengths of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is also the number of minimum (n-)distinguishing labelings of the star graph S_{n+1} on n+1 nodes. - Eric W. Weisstein, Oct 14 2014
When the numbers denote finite permutations (as row numbers of A055089) these are the circular shifts to the right, i.e., a(n) is the permutation with the cycle notation (0 1 ... n-1 n). Compare array A051683 for circular shifts to the right in a broader sense. Compare sequence A007489 for circular shifts to the left. - Tilman Piesk, Apr 29 2017
a(n-1) is the number of permutations on n elements with no cycles of length n. - Dennis P. Walsh, Oct 02 2017
The number of pandigital numbers in base n+1, such that each digit appears exactly once. For example, there are a(9) = 9*9! = 3265920 pandigital numbers in base 10 (A050278). - Amiram Eldar, Apr 13 2020

Examples

			E_1(x) + gamma + log(x) = x/1 - x^2/4 + x^3/18 - x^4/96 + ..., x > 0. - _Michael Somos_, Dec 11 2002
G.f. = x + 4*x^2 + 18*x^3 + 96*x^4 + 600*x^5 + 4320*x^6 + 35280*x^7 + 322560*x^8 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 218.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 336.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 37, equation 37:6:1 at page 354.

Links

Crossrefs

Cf. A000142, A002024, A047920, A047922, A053495, A055089, A123513, A143946, A229837, A239069.
Cf. A163931 (E(x,m,n)), A002775 (n^2*n!), A091363 (n^3*n!), A091364 (n^4*n!).
Cf. sequences with formula (n + k)*n! listed in A282466.
Row sums of A185105, A322383, A322384, A094485.

Programs

  • GAP
    List([0..20], n-> n*Factorial(n) ); # G. C. Greubel, Dec 30 2019
  • Haskell
    a001563 n = a001563_list !! n
a001563_list = zipWith (-) (tail a000142_list) a000142_list
-- Reinhard Zumkeller, Aug 05 2013
  • Magma
    [Factorial(n+1)-Factorial(n): n in [0..20]]; // Vincenzo Librandi, Aug 08 2014
  • Maple
    A001563 := n->n*n!;
  • Mathematica
    Table[n!n,{n,0,25}] (* Harvey P. Dale, Oct 03 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n * n!)} /* Michael Somos, Dec 11 2002 */
  • Sage
    [n*factorial(n) for n in (0..20)] # G. C. Greubel, Dec 30 2019

Formula

From Michael Somos, Dec 11 2002: (Start)
E.g.f.: x / (1 - x)^2.
a(n) = -A021009(n, 1), n >= 0. (End)
The coefficient of y^(n-1) in expansion of (y+n!)^n, n >= 1, gives the sequence 1, 4, 18, 96, 600, 4320, 35280, ... - Artur Jasinski, Oct 22 2007
Integral representation as n-th moment of a function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*(x*(x-1)*exp(-x)) dx, for n>=0. This representation may not be unique. - Karol A. Penson, Sep 27 2001
a(0)=0, a(n) = n*a(n-1) + n!. - Benoit Cloitre, Feb 16 2003
a(0) = 0, a(n) = (n - 1) * (1 + Sum_{i=1..n-1} a(i)) for i > 0. - Gerald McGarvey, Jun 11 2004
Arises in the denominators of the following identities: Sum_{n>=1} 1/(n*(n+1)*(n+2)) = 1/4, Sum_{n>=1} 1/(n*(n+1)*(n+2)*(n+3)) = 1/18, Sum_{n>=1} 1/(n*(n+1)*(n+2)*(n+3)*(n+4)) = 1/96, etc. The general expression is Sum_{n>=k} 1/C(n, k) = k/(k-1). - Dick Boland, Jun 06 2005 [And the general expression implies that Sum_{n>=1} 1/(n*(n+1)*...*(n+k-1)) = (Sum_{n>=k} 1/C(n, k))/k! = 1/((k-1)*(k-1)!) = 1/a(k-1), k >= 2. - Jianing Song, May 07 2023]
a(n) = Sum_{m=2..n+1} |Stirling1(n+1, m)|, n >= 1 and a(0):=0, where Stirling1(n, m) = A048994(n, m), n >= m = 0.
a(n) = 1/(Sum_{k>=0} k!/(n+k+1)!), n > 0. - Vladeta Jovovic, Sep 13 2006
a(n) = Sum_{k=1..n(n+1)/2} k*A143946(n,k). - Emeric Deutsch, Sep 21 2008
The reciprocals of a(n) are the lead coefficients in the factored form of the polynomials obtained by summing the binomial coefficients with a fixed lower term up to n as the upper term, divided by the term index, for n >= 1: Sum_{k = i..n} C(k, i)/k = (1/a(n))*n*(n-1)*..*(n-i+1). The first few such polynomials are Sum_{k = 1..n} C(k, 1)/k = (1/1)*n, Sum_{k = 2..n} C(k, 2)/k = (1/4)*n*(n-1), Sum_{k = 3..n} C(k, 3)/k = (1/18)*n*(n-1)*(n-2), Sum_{k = 4..n} C(k, 4)/k = (1/96)*n*(n-1)*(n-2)*(n-3), etc. - Peter Breznay (breznayp(AT)uwgb.edu), Sep 28 2008
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)* Stirling2(j,i)*x^(k-j) then a(n) = (-1)^(n-1)*f(n,1,-2), (n >= 1). - Milan Janjic, Mar 01 2009
Sum_{n>=1} (-1)^(n+1)/a(n) = 0.796599599... [Jolley eq. 289]
G.f.: 2*x*Q(0), where Q(k) = 1 - 1/(k+2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3)-1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 19 2013
G.f.: W(0)*(1-sqrt(x)) - 1, where W(k) = 1 + sqrt(x)/( 1 - sqrt(x)*(k+2)/(sqrt(x)*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 18 2013
G.f.: T(0)/x - 1/x, where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013
G.f.: Q(0)*(1-x)/x - 1/x, where Q(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1)/( x*(k+1) - 1/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013
D-finite with recurrence: a(n) +(-n-2)*a(n-1) +(n-1)*a(n-2)=0. - R. J. Mathar, Jan 14 2020
a(n) = (-1)^(n+1)*(n+1)*Sum_{k=1..n} A094485(n,k)*Bernoulli(k). The inverse of the Worpitzky representation of the Bernoulli numbers. - Peter Luschny, May 28 2020
From Amiram Eldar, Aug 04 2020: (Start)
Sum_{n>=1} 1/a(n) = Ei(1) - gamma = A229837.
Sum_{n>=1} (-1)^(n+1)/a(n) = gamma - Ei(-1) = A239069. (End)
a(n) = Gamma(n)*A000290(n) for n > 0. - Jacob Szlachetka, Jan 01 2022

A003602 Kimberling's paraphrases: if n = (2k-1)*2^m then a(n) = k.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 17, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22, 6, 23, 12, 24, 2, 25, 13, 26, 7, 27, 14, 28, 4, 29, 15, 30, 8, 31, 16, 32, 1, 33, 17, 34, 9, 35, 18, 36, 5, 37, 19, 38, 10, 39, 20, 40, 3, 41, 21, 42
Offset: 1

Views

Author

N. J. A. Sloane, Mira Bernstein

Keywords

Comments

Fractal sequence obtained from powers of 2.
k occurs at (2*k-1)*A000079(m), m >= 0. - Robert G. Wilson v, May 23 2006
Sequence is T^(oo)(1) where T is acting on a word w = w(1)w(2)..w(m) as follows: T(w) = "1"w(1)"2"w(2)"3"(...)"m"w(m)"m+1". For instance T(ab) = 1a2b3. Thus T(1) = 112, T(T(1)) = 1121324, T(T(T(1))) = 112132415362748. - Benoit Cloitre, Mar 02 2009
Note that iterating the post-numbering operator U(w) = w(1) 1 w(2) 2 w(3) 3... produces the same limit sequence except with an additional "1" prepended, i.e., 1,1,1,2,1,3,2,4,... - Glen Whitney, Aug 30 2023
In the binary expansion of n, first swallow all zeros from the right, then add 1, and swallow the now-appearing 0 bit as well. - Ralf Stephan, Aug 22 2013
Although A264646 and this sequence initially agree in their digit-streams, they differ after 48 digits. - N. J. A. Sloane, Nov 20 2015
"[This is a] fractal because we get the same sequence after we delete from it the first appearance of all positive integers" - see Cobeli and Zaharescu link. - Robert G. Wilson v, Jun 03 2018
From Peter Munn, Jun 16 2022: (Start)
The sequence is the list of positive integers interleaved with the sequence itself. Provided the offset is suitable (which is the case here) a term of such a self-interleaved sequence is determined by the odd part of its index. Putting some of the formulas given here into words, a(n) is the position of the odd part of n in the list of odd numbers.
Applying the interleaving transform again, we get A110963.
(End)
Omitting all 1's leaves A131987 + 1. - David James Sycamore, Jul 26 2022
a(n) is also the smallest positive number not among the terms between a(a(n-1)) and a(n-1) inclusive (with a(0)=1 prepended). - Neal Gersh Tolunsky, Mar 07 2023

Examples

			From _Peter Munn_, Jun 14 2022: (Start)
Start of table showing the interleaving with the positive integers:
   n  a(n)  (n+1)/2  a(n/2)
   1    1      1
   2    1               1
   3    2      2
   4    1               1
   5    3      3
   6    2               2
   7    4      4
   8    1               1
   9    5      5
  10    3               3
  11    6      6
  12    2               2
(End)

References

  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

a(n) is the index of the column in A135764 where n appears (see also A054582).
Cf. A000079, A000265, A001511, A003603, A003961, A014577 (with offset 1, reduction mod 2), A025480, A035528, A048673, A101279, A110963, A117303, A126760, A181988, A220466, A249745, A253887, A337821 (2-adic valuation).
Cf. also A349134 (Dirichlet inverse), A349135 (sum with it), A349136 (Möbius transform), A349431, A349371 (inverse Möbius transform).
Cf. A065120, A131987.
Cf. A264646.
Cf. A123390, A118319.

Programs

  • Haskell
    a003602 = (`div` 2) . (+ 1) . a000265
-- Reinhard Zumkeller, Feb 16 2012, Oct 14 2010
  • Haskell
    import Data.List (transpose)
a003602 = flip div 2 . (+ 1) . a000265
a003602_list = concat $ transpose [[1..], a003602_list]
-- Reinhard Zumkeller, Aug 09 2013, May 23 2013
  • Maple
    A003602:=proc(n) options remember: if n mod 2 = 1 then RETURN((n+1)/2) else RETURN(procname(n/2)) fi: end proc:
seq(A003602(n), n=1..83); # Pab Ter
nmax := 83: for m from 0 to ceil(simplify(log[2](nmax))) do for k from 1 to ceil(nmax/(m+2)) do a((2*k-1)*2^m) := k od: od: seq(a(k), k=1..nmax); # Johannes W. Meijer, Feb 04 2013
A003602 := proc(n)
    a := 1;
    for p in ifactors(n)[2] do
        if op(1,p) > 2 then
            a := a*op(1,p)^op(2,p) ;
        end if;
    end do  :
    (a+1)/2 ;
end proc: # R. J. Mathar, May 19 2016
  • Mathematica
    a[n_] := Block[{m = n}, While[ EvenQ@m, m /= 2]; (m + 1)/2]; Array[a, 84] (* or *)
a[1] = 1; a[n_] := a[n] = If[OddQ@n, (n + 1)/2, a[n/2]]; Array[a, 84] (* Robert G. Wilson v, May 23 2006 *)
a[n_] := Ceiling[NestWhile[Floor[#/2] &, n, EvenQ]/2]; Array[a, 84] (* Birkas Gyorgy, Apr 05 2011 *)
a003602 = {1}; max = 7; Do[b = {}; Do[AppendTo[b, {k, a003602[[k]]}], {k, Length[a003602]}]; a003602 = Flatten[b], {n, 2, max}]; a003602 (* L. Edson Jeffery, Nov 21 2015 *)
  • PARI
    A003602(n)=(n/2^valuation(n,2)+1)/2; /* Joerg Arndt, Apr 06 2011 */
  • Python
    import math
def a(n): return (n/2**int(math.log(n - (n & n - 1), 2)) + 1)/2 # Indranil Ghosh, Apr 24 2017
  • Python
    def A003602(n): return (n>>(n&-n).bit_length())+1 # Chai Wah Wu, Jul 08 2022
  • Scheme
    (define (A003602 n) (let loop ((n n)) (if (even? n) (loop (/ n 2)) (/ (+ 1 n) 2)))) ;; Antti Karttunen, Feb 04 2015

Formula

a(n) = (A000265(n) + 1)/2.
a((2*k-1)*2^m) = k, for m >= 0 and k >= 1. - Robert G. Wilson v, May 23 2006
Inverse Weigh transform of A035528. - Christian G. Bower
G.f.: 1/x * Sum_{k>=0} x^2^k/(1-2*x^2^(k+1) + x^2^(k+2)). - Ralf Stephan, Jul 24 2003
a(2*n-1) = n and a(2*n) = a(n). - Pab Ter (pabrlos2(AT)yahoo.com), Oct 25 2005
a(A118413(n,k)) = A002024(n,k); = a(A118416(n,k)) = A002260(n,k); a(A014480(n)) = A001511(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A001511. - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A249745(A126760(A003961(n))) = A249745(A253887(A048673(n))). That is, this sequence plays the same role for the numbers in array A135764 as A126760 does for the odd numbers in array A135765. - Antti Karttunen, Feb 04 2015 & Jan 19 2016
G.f. satisfies g(x) = g(x^2) + x/(1-x^2)^2. - Robert Israel, Apr 24 2015
a(n) = A181988(n)/A001511(n). - L. Edson Jeffery, Nov 21 2015
a(n) = A025480(n-1) + 1. - R. J. Mathar, May 19 2016
a(n) = A110963(2n-1) = A349135(4*n). - Antti Karttunen, Apr 18 2022
a(n) = (1 + n)/2, for n odd; a(n) = a(n/2), for n even. - David James Sycamore, Jul 28 2022
a(n) = n/2^A001511(n) + 1/2. - Alan Michael Gómez Calderón, Oct 06 2023
a(n) = A123390(A118319(n)). - Flávio V. Fernandes, Mar 02 2025

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Oct 25 2005

A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.

Original entry on oeis.org

0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040, 7199369738058939, 41961001862379596, 244566641436218639
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).
Members of Diophantine pairs. Solution to a*(a+1) = 2*b*(b+1) in natural numbers including 0; a = a(n), b = b(n) = A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
The three sequences x (A001652), y (A046090) and z (A001653) may be obtained by setting u and v equal to the Pell numbers (A000129) in the formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2 [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003
All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006
Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - Kenneth J Ramsey, Sep 22 2007
In general, if b(n) = A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. - Charlie Marion, Nov 19 2007
Limit_{n -> oo} a(n)/a(n-1) = 3+2*sqrt(2) = A156035. - Klaus Brockhaus, Feb 17 2009
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with pMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with pMohamed Bouhamida, Sep 02 2009
a(n+k) = A001541(k)*a(n) + A001542(k)*A001653(n+1) + A001108(k). - Charlie Marion, Dec 10 2010
The numbers 3*A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - Bruno Berselli, Aug 26 2010
Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
T(a(n) + 2*k*A001653(n+1)) = 2*T(A053141(n-1) + k*A002315(n)) + k^2 and
T(a(n) + (2*k+1)*A001653(n+1)) = (A001109(n+1) + k*A002315(n))^2 + k*(k+1).
Also (a(n) + k*A001653(n))^2 + (a(n) + k*A001653(n) + 1)^2 = (A001653(n+1) + k*A002315(n))^2 + k^2. - Charlie Marion, Dec 09 2010
For n>0, A143608(n) divides a(n). - Kenneth J Ramsey, Jun 28 2012
Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 2013
The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014
Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016
Numbers m such that m^2+(m+1)^2 is a square. - César Aguilera, Aug 14 2017
For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018
The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019
Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - Steven Blasberg, Mar 04 2021

Examples

			The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...

References

  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A046090(n) = -a(-1-n).
Cf. A001108, A143608, A089950 (partial sums), A156035.
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; this sequence for k=4; A129556 for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013
Cf. A053141, A001541, A001109, A274307.
Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1).
Cf. A000129, A001333, A001542, A002024, A002378, A029549, A048739, A053142, A055997, A075841, A084159, A084703, A097571, A123737.

Programs

  • GAP
    a:=[0,3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # Muniru A Asiru, Dec 08 2018
  • Haskell
    a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
-- Reinhard Zumkeller, Jan 10 2012
  • Magma
    Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Feb 17 2009
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018
  • Maple
    A001652 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,3]) ;
    else
        6*procname(n-1)-procname(n-2)+2 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016
  • Mathematica
    LinearRecurrence[{7,-7,1}, {0,3,20}, 30] (* Harvey P. Dale, Aug 19 2011 *)
With[{c=3+2*Sqrt[2]},NestList[Floor[c*#]+3&,3,30]] (* Harvey P. Dale, Oct 22 2012 *)
CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *)
Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *)
  • PARI
    {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */
  • PARI
    concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015
  • PARI
    {a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018
  • Sage
    (x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019

Formula

G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n. - Antonio Alberto Olivares, Oct 13 2003
a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005
a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010
Binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). See A053141. - Bill Gosper, Feb 07 2010
Let b(n) = A046090(n) and c(n) = A001653(n). Then for k>j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2*k+1) + b(n)*b(n+2*k+1) + c(n)*c(n+2*k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2*k) + b(n)*b(n+2*k) + c(n)*c(n+2*k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003
For n and j >= 1, Sum_{k=0..j} A001653(k)*a(n) - Sum_{k=0...j-1} A001653(k)*a(n-1) + A053141(j) = A001109(j+1)*a(n) - A001109(j)*a(n-1) + A053141(j) = a(n+j). - Charlie Marion, Jul 07 2003
Sum_{k=0...n} (2*k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - Charlie Marion, Jul 18 2003
a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - Charlie Marion, Jul 21 2003
a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003
a(2*n+k) + a(k) + 1 = A001541(n)*A002315(n+k). For k>0, a(2*n+k) - a(k-1) = A001541(n+k)*A002315(n); e.g., 803760-119 = 19601*41. - Charlie Marion, Mar 17 2003
a(n) = (A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy, Jul 13 2004
a(n) = {2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy, Jul 21 2004
a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8*a(n) +4) + 1, a(1)=0. - Richard Choulet, Sep 18 2007
As noted (Sep 20 2006), a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. In general, for n > 2*k, a(n) = A001653(k)*(a(n-k) + a(n-k-1) + 1) - a(n-2*k-1) - 1. Also a(n) = 7*(a(n-1) - a(n-2)) + a(n-3). In general, for n > 2*k, A002378(k)*(a(n-k)-a(n-k-1)) + a(n-2*k-1). - Charlie Marion, Dec 26 2007
In general, for n >= k >0, a(n) = (A001653(n+k) - A001541(k) * A001653(n) - 2*A001109(k-1))/(4*A001109(k-1)); e.g., 4059 = (33461-3*5741-2*1)/(4*1); 4059 = (195025-17*5741-2*6)/(4*6). - Charlie Marion, Jan 21 2008
From Charlie Marion, Jan 04 2010: (Start)
a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = (A001333(2n+1) - 1)/2.
a(2*n+k-1) = Pell(2*n-1)*Pell(2*n+2*k) + Pell(2*n-2)*Pell(2*n+2*k+1) + A001108(k+1);
a(2*n+k) = Pell(2*n)*Pell(2*n+2*k+1) + Pell(2*n-1)*Pell(2*n+2*k+2) - A055997(k+2). (End)
a(n) = A048739(2*n-1) for n > 0. - Richard R. Forberg, Aug 31 2013
a(n+1) = 3*a(n) + 2*A001653(n) + 1 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - Pierre CAMI, Mar 30 2005; clarified by Hermann Stamm-Wilbrandt, Aug 31 2014
a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - Hermann Stamm-Wilbrandt, Sep 09 2014
For n>0, a(n) = Sum_{k=1..2*n} A000129(k). - Charlie Marion, Nov 07 2015
a(n) = 3*A053142(n) - A053142(n-1). - R. J. Mathar, Feb 05 2016
E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - Ilya Gutkovskiy, Apr 11 2016
a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - Dimitri Papadopoulos, Jul 06 2017
a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - Ezhilarasu Velayutham, Mar 10 2019
a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
In general, for each k >= 0, a(n) = ((a(n-k)+a(k-1)+1)*(a(n-k)-a(k)))/a(n-2*k) for n > 2*k. - Charlie Marion, Dec 27 2020
A generalization of the identity a(n)^2 + A046090(n)^2 = A001653(n+1)^2 follows. Let P(k,n) be the n-th k-gonal number. Then P(k,a(n)) + P(k,A046090(n)) = P(k,A001653(n+1)) + (4-k)*A001109(n). - Charlie Marion, Dec 07 2021
a(n) = A046090(n)-1 = A002024(A029549(n)). - Pontus von Brömssen, Sep 11 2024

Extensions

Additional comments from Wolfdieter Lang, Feb 10 2000

A023758 Numbers of the form 2^i - 2^j with i >= j.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 96, 112, 120, 124, 126, 127, 128, 192, 224, 240, 248, 252, 254, 255, 256, 384, 448, 480, 496, 504, 508, 510, 511, 512, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1023
Offset: 1

Views

Author

Olivier Gérard

Keywords

Comments

Numbers whose digits in base 2 are in nonincreasing order.
Might be called "nialpdromes".
Subset of A077436. Proof: Since a(n) is of the form (2^i-1)*2^j, i,j >= 0, a(n)^2 = (2^(2i) - 2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004
Numbers whose binary representation contains no "01". - Benoit Cloitre, May 23 2004
Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008
From Gary W. Adamson, Jul 18 2008: (Start)
As a triangle by rows starting:
1;
2, 3;
4, 6, 7;
8, 12, 14, 15;
16, 24, 28, 30, 31;
...,
equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8; ...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129, ...). (End)
First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1; ... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009
Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013
From Andres Cicuttin, Apr 29 2016: (Start)
Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).
Examples: for a(n) represented by three bits
Binary
a(5)= 4 -> 100 last bit = 0
a(6)= 6 -> 110 first bit = 1 (inverted last bit of previous number)
a(7)= 7 -> 111
and for a(n) represented by four bits
Binary
a(8) = 8 -> 1000
a(9) = 12 -> 1100 last bit = 0
a(10)= 14 -> 1110 first bit = 1 (inverted last bit of previous number)
a(11)= 15 -> 1111
(End)
Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017
Numbers k such that k = 2^(1 + A000523(k)) - 2^A007814(k). - Daniel Starodubtsev, Aug 05 2021
A002260(n) = v(a(n)/2^v(a(n))+1) and A002024(n) = A002260(n) + v(a(n)) where v is the dyadic valuation (i.e., A007814). - Lorenzo Sauras Altuzarra, Feb 01 2023

Examples

			a(22) = 64 = 32 + 32 = 2^5 + a(16) = 2^A003056(20) + a(22-5-1).
a(23) = 96 = 64 + 32 = 2^6 + a(16) = 2^A003056(21) + a(23-6-1).
a(24) = 112 = 64 + 48 = 2^6 + a(17) = 2^A003056(22) + a(24-6-1).

Links

Crossrefs

A000337(r) = sum of row T(r, c) with 0 <= c < r. See also A002024, A003056, A140129, A140130, A221975.
Cf. A007088, A130123, A101082 (complement), A340375 (characteristic function).
This is the base-2 version of A064222. First differences are A057728.
Subsequence of A077436, of A129523, of A277704, and of A333762.
Subsequences: A043569 (nonzero even terms, or equally, nonzero terms doubled), A175332, A272615, A335431, A000396 (its even terms only), A324200.
Positions of zeros in A049502, A265397, A277899, A284264.
Positions of ones in A283983, A283989.
Positions of nonzero terms in A341509 (apart from the initial zero).
Positions of squarefree terms in A260443.
Fixed points of A264977, A277711, A283165, A334666.
Distinct terms in A340632.
Cf. also A002024, A002260, A007814, A133797, A152449, A181666, A211705, A277699, A306441.
Cf. also A309758, A309759, A309761 (for analogous sequences).

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
a023758 n = a023758_list !! (n-1)
a023758_list = 0 : f (singleton 1) where
f s = x : f (if even x then insert z s' else insert z $ insert (z+1) s')
where z = 2*x; (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 24 2014, Dec 19 2012
  • Maple
    a:=proc(n) local n2,d: n2:=convert(n,base,2): d:={seq(n2[j]-n2[j-1],j=2..nops(n2))}: if n=0 then 0 elif n=1 then 1 elif d={0,1} or d={0} or d={1} then n else fi end: seq(a(n),n=0..2100); # Emeric Deutsch, Apr 22 2006
  • Mathematica
    Union[Flatten[Table[2^i - 2^j, {i, 0, 100}, {j, 0, i}]]] (* T. D. Noe, Mar 15 2011 *)
Select[Range[0, 2^10], NoneTrue[Differences@ IntegerDigits[#, 2], # > 0 &] &] (* Michael De Vlieger, Sep 05 2017 *)
  • PARI
    for(n=0,2500,if(prod(k=1,length(binary(n))-1,component(binary(n),k)+1-component(binary(n),k+1))>0,print1(n,",")))
  • PARI
    A023758(n)= my(r=round(sqrt(2*n--))); (1<<(n-r*(r-1)/2)-1)<<(r*(r+1)/2-n)
/* or, to illustrate the "decreasing digit" property and analogy to A064222: */
A023758(n,show=0)={ my(a=0); while(n--, show & print1(a","); a=vecsort(binary(a+1)); a*=vector(#a,j,2^(j-1))~); a} \\ M. F. Hasler, May 06 2009
  • PARI
    is(n)=if(n<5,1,n>>=valuation(n,2);n++;n>>valuation(n,2)==1) \\ Charles R Greathouse IV, Jan 04 2016
  • PARI
    list(lim)=my(v=List([0]),t); for(i=1,logint(lim\1+1,2), t=2^i-1; while(t<=lim, listput(v,t); t*=2)); Set(v) \\ Charles R Greathouse IV, May 03 2016
  • Python
    def a_next(a_n): return (a_n | (a_n >> 1)) + (a_n & 1)
a_n = 1; a = [0]
for i in range(55): a.append(a_n); a_n = a_next(a_n) # Falk Hüffner, Feb 19 2022
  • Python
    from math import isqrt
def A023758(n): return (1<<(m:=isqrt(n-1<<3)+1>>1))-(1<<(m*(m+1)-(n-1<<1)>>1)) # Chai Wah Wu, Feb 23 2025

Formula

a(n) = 2^s(n) - 2^((s(n)^2 + s(n) - 2n)/2) where s(n) = ceiling((-1 + sqrt(1+8n))/2). - Sam Alexander, Jan 08 2005
a(n) = 2^k + a(n-k-1) for 1 < n and k = A003056(n-2). The rows of T(r, c) = 2^r-2^c for 0 <= c < r read from right to left produce this sequence: 1; 2, 3; 4, 6, 7; 8, 12, 14, 15; ... - Frank Ellermann, Dec 06 2001
For n > 0, a(n) mod 2 = A010054(n). - Benoit Cloitre, May 23 2004
A140130(a(n)) = 1 and for n > 1: A140129(a(n)) = A002262(n-2). - Reinhard Zumkeller, May 14 2008
a(n+1) = (2^(n - r(r-1)/2) - 1) 2^(r(r+1)/2 - n), where r=round(sqrt(2n)). - M. F. Hasler, May 06 2009
Start with A000225. If k is in the sequence, then so is 2k. - Ralf Stephan, Aug 16 2013
G.f.: (x^2/((2-x)*(1-x)))*(1 + Sum_{k>=0} x^((k^2+k)/2)*(1 + x*(2^k-1))). The sum is related to Jacobi theta functions. - Robert Israel, Feb 24 2015
A049502(a(n)) = 0. - Reinhard Zumkeller, Jun 17 2015
a(n) = a(n-1) + a(n-d)/a(d*(d+1)/2 + 2) if n > 1, d > 0, where d = A002262(n-2). - Yuchun Ji, May 11 2020
A277699(a(n)) = a(n)^2, A306441(a(n)) = a(n+1). - Antti Karttunen, Feb 15 2021 (the latter identity from A306441)
Sum_{n>=2} 1/a(n) = A211705. - Amiram Eldar, Feb 20 2022

Extensions

Definition changed by N. J. A. Sloane, Jan 05 2008

A018900 Sums of two distinct powers of 2.

Original entry on oeis.org

3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 65, 66, 68, 72, 80, 96, 129, 130, 132, 136, 144, 160, 192, 257, 258, 260, 264, 272, 288, 320, 384, 513, 514, 516, 520, 528, 544, 576, 640, 768, 1025, 1026, 1028, 1032, 1040, 1056, 1088, 1152, 1280, 1536, 2049, 2050, 2052, 2056, 2064, 2080, 2112, 2176, 2304, 2560, 3072
Offset: 1

Views

Author

Jonn Dalton (jdalton(AT)vnet.ibm.com), Dec 11 1996

Keywords

Comments

Appears to give all k such that 8 is the highest power of 2 dividing A005148(k). - Benoit Cloitre, Jun 22 2002
Seen as a triangle read by rows, T(n,k) = 2^(k-1) + 2^n, 1 <= k <= n, the sum of the n-th row equals A087323(n). - Reinhard Zumkeller, Jun 24 2009
Numbers whose base-2 sum of digits is 2. - Tom Edgar, Aug 31 2013
All odd terms are A000051. - Robert G. Wilson v, Jan 03 2014
A239708 holds the subsequence of terms m such that m - 1 is prime. - Hieronymus Fischer, Apr 20 2014

Examples

			From _Hieronymus Fischer_, Apr 27 2014: (Start)
a(1) = 3, since 3 = 2^1 + 2^0.
a(5) = 10, since 10 = 2^3 + 2^1.
a(10^2) = 16640
a(10^3) = 35184372089344
a(10^4) = 2788273714550169769618891533295908724670464 = 2.788273714550...*10^42
a(10^5) = 3.6341936214780344527466190...*10^134
a(10^6) = 4.5332938264998904048012398...*10^425
a(10^7) = 1.6074616084721302346802429...*10^1346
a(10^8) = 1.4662184497310967196301632...*10^4257
a(10^9) = 2.3037539289782230932863807...*10^13462
a(10^10) = 9.1836811272250798973464436...*10^42571
(End)

Links

Crossrefs

Cf. A000120, A001969, A048639, A048645, A057168.
Cf. A000217, A179951, A187813, A239708.
Cf. A000079, A014311, A014312, A014313, A023688, A023689, A023690, A023691 (Hamming weight = 1, 3, 4, ..., 9).
Sum of base-b digits equal b: A226636 (b = 3), A226969 (b = 4), A227062 (b = 5), A227080 (b = 6), A227092 (b = 7), A227095 (b = 8), A227238 (b = 9), A052224 (b = 10). - M. F. Hasler, Dec 23 2016

Programs

  • C
    unsigned hakmem175(unsigned x) {
    unsigned s, o, r;
    s = x & -x; r = x + s;
    o = x ^ r;  o = (o >> 2) / s;
    return r | o;
}
unsigned A018900(int n) {
    if (n == 1) return 3;
    return hakmem175(A018900(n - 1));
} // Peter Luschny, Jan 01 2014
  • Haskell
    a018900 n = a018900_list !! (n-1)
a018900_list = elemIndices 2 a073267_list  -- Reinhard Zumkeller, Mar 07 2012
  • Maple
    a:= n-> (i-> 2^i+2^(n-1-i*(i-1)/2))(floor((sqrt(8*n-1)+1)/2)):
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 01 2022
  • Mathematica
    Select[ Range[ 1056 ], (Count[ IntegerDigits[ #, 2 ], 1 ]==2)& ]
Union[Total/@Subsets[2^Range[0,10],{2}]] (* Harvey P. Dale, Mar 04 2012 *)
  • PARI
    for(m=1,9,for(n=0,m-1,print1(2^m+2^n", "))) \\ Charles R Greathouse IV, Sep 09 2011
  • PARI
    is(n)=hammingweight(n)==2 \\ Charles R Greathouse IV, Mar 03 2014
  • PARI
    for(n=0,10^5,if(hammingweight(n)==2,print1(n,", "))); \\ Joerg Arndt, Mar 04 2014
  • PARI
    a(n)= my(t=sqrtint(n*8)\/2); 2^t + 2^(n-1-t*(t-1)/2); \\ Ruud H.G. van Tol, Nov 30 2024
  • Python
    print([n for n in range(1, 3001) if bin(n)[2:].count("1")==2]) # Indranil Ghosh, Jun 03 2017
  • Python
    A018900_list = [2**a+2**b for a in range(1,10) for b in range(a)] # Chai Wah Wu, Jan 24 2021
  • Python
    from math import isqrt, comb
def A018900(n): return (1<<(m:=isqrt(n<<3)+1>>1))+(1<<(n-1-comb(m,2))) # Chai Wah Wu, Oct 30 2024
  • Smalltalk
    distinctPowersOf: b
  "Version 1: Answers the n-th number of the form b^i + b^j, i>j>=0, where n is the receiver.
  b > 1 (b = 2, for this sequence).
  Usage: n distinctPowersOf: 2
  Answer: a(n)"
  | n i j |
  n := self.
  i := (8*n - 1) sqrtTruncated + 1 // 2.
  j := n - (i*(i - 1)/2) - 1.
  ^(b raisedToInteger: i) + (b raisedToInteger: j)
[by Hieronymus Fischer, Apr 20 2014]
------------
  • Smalltalk
    distinctPowersOf: b
  "Version 2: Answers an array which holds the first n numbers of the form b^i + b^j, i>j>=0, where n is the receiver. b > 1 (b = 2, for this sequence).
  Usage: n distinctPowersOf: 2
  Answer: #(3 5 6 9 10 12 ...) [first n terms]"
  | k p q terms |
  terms := OrderedCollection new.
  k := 0.
  p := b.
  q := 1.
  [k < self] whileTrue:
         [[q < p and: [k < self]] whileTrue:
                   [k := k + 1.
                   terms add: p + q.
                   q := b * q].
         p := b * p.
         q := 1].
  ^terms as Array
[by Hieronymus Fischer, Apr 20 2014]
------------
  • Smalltalk
    floorDistinctPowersOf: b
  "Answers an array which holds all the numbers b^i + b^j < n, i>j>=0, where n is the receiver.
  b > 1 (b = 2, for this sequence).
  Usage: n floorDistinctPowersOf: 2
  Answer: #(3 5 6 9 10 12 ...) [all terms < n]"
  | a n p q terms |
  terms := OrderedCollection new.
  n := self.
  p := b.
  q := 1.
  a := p + q.
  [a < n] whileTrue:
         [[q < p and: [a < n]] whileTrue:
                   [terms add: a.
                   q := b * q.
                   a := p + q].
         p := b * p.
         q := 1.
         a := p + q].
  ^terms as Array
[by Hieronymus Fischer, Apr 20 2014]
------------
  • Smalltalk
    invertedDistinctPowersOf: b
  "Given a number m which is a distinct power of b, this method answers the index n such that there are uniquely defined i>j>=0 for which b^i + b^j = m, where m is the receiver;  b > 1 (b = 2, for this sequence).
  Usage: m invertedDistinctPowersOf: 2
  Answer: n such that a(n) = m, or, if no such n exists, min (k | a(k) >= m)"
  | n i j k m |
  m := self.
  i := m integerFloorLog: b.
  j := m - (b raisedToInteger: i) integerFloorLog: b.
  n := i * (i - 1) / 2 + 1 + j.
  ^n
[by Hieronymus Fischer, Apr 20 2014]

Formula

a(n) = 2^trinv(n-1) + 2^((n-1)-((trinv(n-1)*(trinv(n-1)-1))/2)), i.e., 2^A002024(n)+2^A002262(n-1). - Antti Karttunen
a(n) = A059268(n-1) + A140513(n-1). A000120(a(n)) = 2. Complement of A161989. A151774(a(n)) = 1. - Reinhard Zumkeller, Jun 24 2009
A073267(a(n)) = 2. - Reinhard Zumkeller, Mar 07 2012
Start with A000051. If n is in sequence, then so is 2n. - Ralf Stephan, Aug 16 2013
a(n) = A057168(a(n-1)) for n>1 and a(1) = 3. - Marc LeBrun, Jan 01 2014
From Hieronymus Fischer, Apr 20 2014: (Start)
Formulas for a general parameter b according to a(n) = b^i + b^j, i>j>=0; b = 2 for this sequence.
a(n) = b^i + b^j, where i = floor((sqrt(8n - 1) + 1)/2), j = n - 1 - i*(i - 1)/2 [for a Smalltalk implementation see Prog section, method distinctPowersOf: b (2 versions)].
a(A000217(n)) = (b + 1)*b^(n-1) = b^n + b^(n-1).
a(A000217(n)+1) = 1 + b^(n+1).
a(n + 1 + floor((sqrt(8n - 1) + 1)/2)) = b*a(n).
a(n + 1 + floor(log_b(a(n)))) = b*a(n).
a(n + 1) = b^2/(b+1) * a(n) + 1, if n is a triangular number (s. A000217).
a(n + 1) = b*a(n) + (1-b)* b^floor((sqrt(8n - 1) + 1)/2), if n is not a triangular number.
The next term can also be calculated without using the index n. Let m be a term and i = floor(log_b(m)), then:
a(n + 1) = b*m + (1-b)* b^i, if floor(log_b(m/(b+1))) + 1 < i,
a(n + 1) = b^2/(b+1) * m + 1, if floor(log_b(m/(b+1))) + 1 = i.
Partial sum:
Sum_{k=1..n} a(k) = ((((b-1)*(j+1)+i-1)*b^(i-j) + b)*b^j - i)/(b-1), where i = floor((sqrt(8*n - 1) + 1)/2), j = n - 1 - i*(i - 1)/2.
Inverse:
For each sequence term m, the index n such that a(n) = m is determined by n := i*(i-1)/2 + j + 1, where i := floor(log_b(m)), j := floor(log_b(m - b^floor(log_b(m)))) [for a Smalltalk implementation see Prog section, method invertedDistinctPowersOf: b].
Inequalities:
a(n) <= (b+1)/b * b^floor(sqrt(2n)+1/2), equality holds for triangular numbers.
a(n) > b^floor(sqrt(2n)+1/2).
a(n) < b^sqrt(2n)*sqrt(b).
a(n) > b^sqrt(2n)/sqrt(b).
Asymptotic behavior:
lim sup a(n)/b^sqrt(2n) = sqrt(b).
lim inf a(n)/b^sqrt(2n) = 1/sqrt(b).
lim sup a(n)/b^(floor(sqrt(2n))) = b.
lim inf a(n)/b^(floor(sqrt(2n))) = 1.
lim sup a(n)/b^(floor(sqrt(2n)+1/2)) = (b+1)/b.
lim inf a(n)/b^(floor(sqrt(2n)+1/2)) = 1.
(End)
Sum_{n>=1} 1/a(n) = A179951. - Amiram Eldar, Oct 06 2020

Extensions

Edited by M. F. Hasler, Dec 23 2016
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