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

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Author

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).

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. 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

A016813 a(n) = 4*n + 1.

Original entry on oeis.org

1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97, 101, 105, 109, 113, 117, 121, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225, 229, 233, 237
Offset: 0

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 23 ).
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 64 ).
Numbers k such that k and (k+1) have the same binary digital sum. - Benoit Cloitre, Jun 05 2002
Numbers k such that (1 + sqrt(k))/2 is an algebraic integer. - Alonso del Arte, Jun 04 2012
Numbers k such that 2 is the only prime p that satisfies the relationship p XOR k = p + k. - Brad Clardy, Jul 22 2012
This may also be interpreted as the array T(n,k) = A001844(n+k) + A008586(k) read by antidiagonals:
1, 9, 21, 37, 57, 81, ...
5, 17, 33, 53, 77, 105, ...
13, 29, 49, 73, 101, 133, ...
25, 45, 69, 97, 129, 165, ...
41, 65, 93, 125, 161, 201, ...
61, 89, 121, 157, 197, 241, ...
...
- R. J. Mathar, Jul 10 2013
With leading term 2 instead of 1, 1/a(n) is the largest tolerance of form 1/k, where k is a positive integer, so that the nearest integer to (n - 1/k)^2 and to (n + 1/k)^2 is n^2. In other words, if interval arithmetic is used to square [n - 1/k, n + 1/k], every value in the resulting interval of length 4n/k rounds to n^2 if and only if k >= a(n). - Rick L. Shepherd, Jan 20 2014
Odd numbers for which the number of prime factors congruent to 3 (mod 4) is even. - Daniel Forgues, Sep 20 2014
For the Collatz conjecture, we identify two types of odd numbers. This sequence contains all the descenders: where (3*a(n) + 1) / 2 is even and requires additional divisions by 2. See A004767 for the ascenders. - Fred Daniel Kline, Nov 29 2014 [corrected by Jaroslav Krizek, Jul 29 2016]
a(n-1), n >= 1, is also the complex dimension of the manifold M(S), the set of all conjugacy classes of irreducible representations of the fundamental group pi_1(X,x_0) of rank 2, where S = {a_1, ..., a_{n}, a_{n+1} = oo}, a subset of P^1 = C U {oo}, X = X(S) = P^1 \ S, and x_0 a base point in X. See the Iwasaki et al. reference, Proposition 2.1.4. p. 150. - Wolfdieter Lang, Apr 22 2016
For n > 3, also the number of (not necessarily maximal) cliques in the n-sunlet graph. - Eric W. Weisstein, Nov 29 2017
For integers k with absolute value in A047202, also exponents of the powers of k having the same unit digit of k in base 10. - Stefano Spezia, Feb 23 2021
Starting with a(1) = 5, numbers ending with 01 in base 2. - John Keith, May 09 2022

Examples

			From _Leo Tavares_, Jul 02 2021: (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
(End)
		

References

  • K. Iwasaki, H. Kimura, S. Shimomura and M. Yoshida, From Gauss to Painlevé, Vieweg, 1991. p. 150.

Crossrefs

Subsequence of A042963 and of A079523.
a(n) = A093561(n+1, 1), (4, 1)-Pascal column.
Cf. A004772 (complement).
Cf. A017557.

Programs

Formula

a(n) = A005408(2*n).
Sum_{n>=0} (-1)^n/a(n) = (1/(4*sqrt(2)))*(Pi+2*log(sqrt(2)+1)) = A181048 [Jolley]. - Benoit Cloitre, Apr 05 2002 [corrected by Amiram Eldar, Jul 30 2023]
G.f.: (1+3*x)/(1-x)^2. - Paul Barry, Feb 27 2003 [corrected for offset 0 by Wolfdieter Lang, Oct 03 2014]
(1 + 5*x + 9*x^2 + 13*x^3 + ...) = (1 + 2*x + 3*x^2 + ...) / (1 - 3*x + 9*x^2 - 27*x^3 + ...). - Gary W. Adamson, Jul 03 2003
a(n) = A001969(n) + A000069(n). - Philippe Deléham, Feb 04 2004
a(n) = A004766(n-1). - R. J. Mathar, Oct 26 2008
a(n) = 2*a(n-1) - a(n-2); a(0)=1, a(1)=5. a(n) = 4 + a(n-1). - Philippe Deléham, Nov 03 2008
A056753(a(n)) = 3. - Reinhard Zumkeller, Aug 23 2009
A179821(a(n)) = a(A179821(n)). - Reinhard Zumkeller, Jul 31 2010
a(n) = 8*n - 2 - a(n-1) for n > 0, a(0) = 1. - Vincenzo Librandi, Nov 20 2010
The identity (4*n+1)^2 - (4*n^2+2*n)*(2)^2 = 1 can be written as a(n)^2 - A002943(n)*2^2 = 1. - Vincenzo Librandi, Mar 11 2009 - Nov 25 2012
A089911(6*a(n)) = 8. - Reinhard Zumkeller, Jul 05 2013
a(n) = A004767(n) - 2. - Jean-Bernard François, Sep 27 2013
a(n) = A058281(3n+1). - Eli Jaffe, Jun 07 2016
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (1 + 4*x)*exp(x).
a(n) = Sum_{k = 0..n} A123932(k).
a(A005098(k)) = x^2 + y^2.
Inverse binomial transform of A014480. (End)
Dirichlet g.f.: 4*Zeta(-1 + s) + Zeta(s). - Stefano Spezia, Nov 02 2018

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

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).

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. A264646.

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

A002457 a(n) = (2n+1)!/n!^2.

Original entry on oeis.org

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900
Offset: 0

Comments

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

Examples

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • 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).
  • J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

Crossrefs

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

Programs

Formula

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k). This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A002420 Expansion of sqrt(1 - 4*x) in powers of x.

Original entry on oeis.org

1, -2, -2, -4, -10, -28, -84, -264, -858, -2860, -9724, -33592, -117572, -416024, -1485800, -5348880, -19389690, -70715340, -259289580, -955277400, -3534526380, -13128240840, -48932534040, -182965127280, -686119227300, -2579808294648, -9723892802904, -36734706144304
Offset: 0

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

Also expansion of complementary modulus k' in powers of m/4 = k^2/4.
Series reversion of x(Sum_{k>=0} a(k)x^(2k)) is x(Sum_{k>=0} C(2k)x^(2k)) where C() is Catalan numbers A000108.
The g.f. of the reciprocal sequence 1,-1/2,-1/2,... is F(1,1;-1/2;x/4). - Paul Barry, Sep 18 2008
Hankel transform is (2n+1)*(-2)^n or (-1)^n*A014480. - Paul Barry, Jan 22 2009
Equals polcoeff inverse of A000984. - Gary W. Adamson, Jun 02 2009
|a(n)| is the number of lattice paths in steps of (1,1) and (1,-1) that begin at the origin and end at (2n,0) but otherwise never touch (or cross) the x axis. Note the paths are in both the first and fourth quadrants. O.g.f. is 2xC(x)+1 where C(x) is the o.g.f. for A000108 (Catalan numbers). - Geoffrey Critzer, Jan 17 2012

Examples

			sqrt(1 - 4*x) = 1 - 2*x - 2*x^2 - 4*x^3 - 10*x^4 - 28*x^5 - 84*x^6 - 264*x^7 - 858*x^8 - 2860*x^9 - ...
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 55.
  • 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).
  • T. N. Thiele, Interpolationsrechnung. Teubner, Leipzig, 1909, p. 164.

Crossrefs

A068875 and A262543 are essentially the same sequence as this.

Programs

  • Magma
    [Binomial(2*n, n)/(1-2*n): n in [0..30]]; // G. C. Greubel, Aug 12 2018
    
  • Maple
    A002420:=n->binomial(2*n, n)/(1-2*n); seq(A002420(n), n=1..30); # Wesley Ivan Hurt, May 08 2014
  • Mathematica
    a[n_] := -2n(2n-2)! / n!^2; a[0] = 1; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Dec 07 2011 *)
    Table[If[n==0,1,-2 CatalanNumber[n-1]], {n,0,27}] (* Peter Luschny, Feb 27 2017 *)
    CoefficientList[Series[Sqrt[1-4x],{x,0,30}],x] (* Harvey P. Dale, Jul 04 2017 *)
  • PARI
    {a(n) = binomial(2*n, n) / (1 - 2*n)} /* Michael Somos, Jul 12 2008 */
    
  • Sage
    [catalan_number(n)*((1+n)/(1-2*n)) for n in range(30)] # G. C. Greubel, Nov 26 2018

Formula

G.f.: sqrt(1-4*x) = 1F0(-1/2;;4*x).
a(n) = binomial(2*n, n)/(1-2*n).
a(n) ~ -(1/2)*Pi^(-1/2)*n^(-3/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
0 = 16 * a(n) * a(k) * a(n+k+1) - 8 * a(n) * a(k) * a(n+k+2) + a(n+1) * a(k) * a(n+k+2) - a(n+1) * a(k+1) * a(n+k+1) + a(n) * a(k+1) * a(n+k+2) for all n and k. - Michael Somos, Jul 12 2008
G.f.: 2F1(1,-1/2;1;4x). - Paul Barry, Jan 22 2009
a(n) = (-1)^n * binomial(1/2,n)*4^n. - Vladimir Kruchinin, May 22 2011
G.f.: A(x) = (1-4*x)^(1/2) = 1 - 2*x - 2*x^2/G(0); G(k) = 1 - 2*x - x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2011
D-finite with recurrence: n*a(n) +2*(3-2*n)*a(n-1)=0. - R. J. Mathar, Dec 19 2011
E.g.f.: a(n) = (-1)^n*n!* [x^n] exp(-2*x)*((1 + 4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). -Peter Luschny, Aug 25 2012
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: 2*G(0) - 1, where G(k) = 2*x*(2*k+1) + (k+1) - 2*x*(k+1)*(2*k+3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 02 2013
a(n) = 4^n * binomial(n-3/2, -3/2). - Peter Luschny, May 06 2014
a(n) = 4^n*hypergeom([-n,3/2],[1],1). - Peter Luschny, Apr 26 2016
From Amiram Eldar, Mar 24 2022: (Start)
Sum_{n>=0} 1/a(n) = -2*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 32/25 - 12*log(phi)/(25*sqrt(5)), where phi is the golden ratio (A001622). (End)
From Peter Bala, Mar 31 2024: (Start)
a(n) = (4^n)*Sum_{k = 0..2*n} (-1)^k*binomial(1/2, k)*binomial(1/2, 2*n - k).
(4^n)*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
(1/2)*Sum_{k = 0..n} a(k)*a(2*n-k) = (Catalan(n-1))^2 = A001246(n) for n >= 1.
Sum_{k = 0..2*n} a(k)*a(2*n-k) = 0 for n >= 1. (End)

Extensions

Additional comments from Michael Somos, Dec 13 2002

A054582 Array read by antidiagonals upwards: A(m,k) = 2^m * (2k+1), m,k >= 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 12, 10, 7, 16, 24, 20, 14, 9, 32, 48, 40, 28, 18, 11, 64, 96, 80, 56, 36, 22, 13, 128, 192, 160, 112, 72, 44, 26, 15, 256, 384, 320, 224, 144, 88, 52, 30, 17, 512, 768, 640, 448, 288, 176, 104, 60, 34, 19, 1024, 1536, 1280, 896, 576, 352, 208, 120
Offset: 0

Author

Henry Bottomley, Apr 12 2000

Keywords

Comments

First column of array is powers of 2, first row is odd numbers, other cells are products of these two, so every positive integer appears exactly once. [Comment edited to match the definition. - L. Edson Jeffery, Jun 05 2015]
An analogous N X N <-> N bijection based, not on the binary, but on the Fibonacci number system, is given by the Wythoff array A035513.
As an array, this sequence (hence also A135764) is the dispersion of the even positive integers. For the definition of dispersion, see the link "Interspersions and Dispersions." The fractal sequence of this dispersion is A003602. - Clark Kimberling, Dec 03 2010

Examples

			Northwest corner of array A:
    1     3     5     7     9    11    13    15    17    19
    2     6    10    14    18    22    26    30    34    38
    4    12    20    28    36    44    52    60    68    76
    8    24    40    56    72    88   104   120   136   152
   16    48    80   112   144   176   208   240   272   304
   32    96   160   224   288   352   416   480   544   608
   64   192   320   448   576   704   832   960  1088  1216
  128   384   640   896  1152  1408  1664  1920  2176  2432
  256   768  1280  1792  2304  2816  3328  3840  4352  4864
  512  1536  2560  3584  4608  5632  6656  7680  8704  9728
[Array edited to match the definition. - _L. Edson Jeffery_, Jun 05 2015]
From _Philippe Deléham_, Dec 13 2013: (Start)
a(13-1)=20=2*10, so a(13)=10+A006519(20)=10+4=14.
a(3-1)=3=2*1+1, so a(3)=2^(1+1)=4. (End)
From _Wolfdieter Lang_, Jan 30 2019: (Start)
The triangle T begins:
   n\k   0    1    2   3   4   5   6   7  8  9 10 ...
   0:    1
   1:    2    3
   2:    4    6    5
   3:    8   12   10   7
   4:   16   24   20  14   9
   5:   32   48   40  28  18  11
   6:   64   96   80  56  36  22  13
   7:  128  192  160 112  72  44  26  15
   8:  256  384  320 224 144  88  52  30 17
   9:  512  768  640 448 288 176 104  60 34 19
  10: 1024 1536 1280 896 576 352 208 120 68 38 21
  ...
T(3, 2) = 2^1*(2*2+1) = 10. (End)
		

Crossrefs

The sequence is a permutation of A000027.
Main diagonal is A014480; inverse permutation is A209268.

Programs

  • Haskell
    a054582 n k = a054582_tabl !! n !! k
    a054582_row n = a054582_tabl !! n
    a054582_tabl = iterate
       (\xs@(x:_) -> (2 * x) : zipWith (+) xs (iterate (`div` 2) (2 * x))) [1]
    a054582_list = concat a054582_tabl
    -- Reinhard Zumkeller, Jan 22 2013
    
  • Mathematica
    (* Array: *)
    Grid[Table[2^m*(2*k + 1), {m, 0, 9}, {k, 0, 9}]] (* L. Edson Jeffery, Jun 05 2015 *)
    (* Array antidiagonals flattened: *)
    Flatten[Table[2^(m - k)*(2*k + 1), {m, 0, 9}, {k, 0, m}]] (* L. Edson Jeffery, Jun 05 2015 *)
  • PARI
    T(m,k)=(2*k+1)<Charles R Greathouse IV, Jun 21 2017

Formula

As a sequence, if n is a triangular number, then a(n)=a(n-A002024(n))+2, otherwise a(n)=2*a(n-A002024(n)-1).
a(n) = A075300(n-1)+1.
Recurrence for the sequence: if a(n-1)=2*k is even, then a(n)=k+A006519(2*k); if a(n-1)=2*k+1 is odd, then a(n)=2^(k+1), a(0)=1. - Philippe Deléham, Dec 13 2013
m = A(A001511(m)-1, A003602(m)-1), for each m in A000027. - L. Edson Jeffery, Nov 22 2015
The triangle is T(n, k) = A(n-k, k) = 2^(n-k)*(2*k+1), for n >= 0 and k = 0..n. - Wolfdieter Lang, Jan 30 2019

Extensions

Offset corrected by Reinhard Zumkeller, Jan 22 2013

A118413 Triangle read by rows: T(n,k) = (2*n-1)*2^(k-1), 0

Original entry on oeis.org

1, 3, 6, 5, 10, 20, 7, 14, 28, 56, 9, 18, 36, 72, 144, 11, 22, 44, 88, 176, 352, 13, 26, 52, 104, 208, 416, 832, 15, 30, 60, 120, 240, 480, 960, 1920, 17, 34, 68, 136, 272, 544, 1088, 2176, 4352, 19, 38, 76, 152, 304, 608, 1216, 2432, 4864, 9728, 21, 42, 84, 168
Offset: 1

Author

Reinhard Zumkeller, Apr 27 2006

Keywords

Comments

Central terms give A118415; row sums give A118414;
T(n,1) = A005408(n-1);
T(n,2) = A016825(n-1) for n>1;
T(n,3) = A017113(n-1) for n>2;
T(n,4) = A051062(n-1) for n>3;
T(n,n-2) = A052951(n-1) for n>2;
T(n,n) = A014480(n-1) = A118416(n,n);
A001511(T(n,k)) = A002260(n,k);
A003602(T(n,k)) = A002024(n,k).
G.f.: x*y*(1 + x + 2*x*y - 6*x^2*y)/((1 - x)^2*(1 - 2*x*y)^2). - Stefano Spezia, Dec 22 2024

Examples

			   1
   3   6
   5  10  20
   7  14  28  56
   9  18  36  72 144
  11  22  44  88 176 352
  13  26  52 104 208 416  832
  15  30  60 120 240 480  960 1920
  17  34  68 136 272 544 1088 2176 4352
  19  38  76 152 304 608 1216 2432 4864 9728
  ...
		

Programs

  • Mathematica
    Select[Flatten[Table[(2n-1)2^(k-1),{n,20},{k,0,n}]],IntegerQ] (* Harvey P. Dale, Jan 17 2024 *)
  • Python
    from math import isqrt, comb
    def A118413(n):
        a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
        return ((a<<1)-1)<Chai Wah Wu, Jun 20 2025

A118416 Triangle read by rows: T(n,k) = (2*k-1)*2^(n-1), 0 < k <= n.

Original entry on oeis.org

1, 2, 6, 4, 12, 20, 8, 24, 40, 56, 16, 48, 80, 112, 144, 32, 96, 160, 224, 288, 352, 64, 192, 320, 448, 576, 704, 832, 128, 384, 640, 896, 1152, 1408, 1664, 1920, 256, 768, 1280, 1792, 2304, 2816, 3328, 3840, 4352, 512, 1536, 2560, 3584, 4608, 5632, 6656, 7680
Offset: 1

Author

Reinhard Zumkeller, Apr 27 2006

Keywords

Comments

Row sums give A014477: Sum_{k=1..n} T(n,k) = A014477(n-1);
central terms give A118415; T(2*k-1,k) = A058962(k-1);
T(n,1) = A000079(n-1);
T(n,2) = A007283(n-1) for n > 1;
T(n,3) = A020714(n-1) for n > 2;
T(n,4) = A005009(n-1) for n > 3;
T(n,5) = A005010(n-1) for n > 4;
T(n,n-1) = A118417(n-1) for n > 1;
T(n,n) = A014480(n-1) = A118413(n,n);
A001511(T(n,k)) = A002024(n,k);
A003602(T(n,k)) = A002260(n,k).
The alternating row sums, Sum_{k=1..n} (-1)^(k+1)*T(n,k), are: (a) in odd rows, the central term, T(n,(n+1)/2) = A058962((n-1)/2); (b) in even rows, the negation of the average of the two central terms, -(T(2n,n) + T(2n,+1))/2 = -A018215(m/2). The absolute values of the alternating row sums give the plain row means, Sum_{k=1..n} T(n,k)/n; the alternating sign row means are (-2)^(n-1). - Gregory Gerard Wojnar, Feb 10 2024

Examples

			Triangle begins:
   1;
   2,   6;
   4,  12,  20;
   8,  24,  40,  56;
  16,  48,  80, 112, 144;
  32,  96, 160, 224, 288, 352;
  64, 192, 320, 448, 576, 704, 832;
		

Crossrefs

Programs

  • Haskell
    a118416 n k = a118416_tabl !! (n-1) !! (k-1)
    a118416_row 1 = [1]
    a118416_row n = (map (* 2) $ a118416_row (n-1)) ++ [a014480 (n-1)]
    a118416_tabl = map a118416_row [1..]
    -- Reinhard Zumkeller, Jan 22 2012
    
  • Maple
    A118416 := proc(n,k) 2^(n-1)*(2*k-1) ; end proc: # R. J. Mathar, Sep 04 2011
  • Mathematica
    Flatten[Table[(2k-1)2^(n-1),{n,10},{k,n}]] (* Harvey P. Dale, Aug 26 2014 *)
  • Python
    from math import isqrt
    def A118416(n): return (a:=(m:=isqrt(k:=n<<1))+(k>m*(m+1)))*(1-a)+(n<<1)-1<Chai Wah Wu, Jun 20 2025

Formula

T(n,k) = 2*T(n-1,k), 1 <= k < n; T(n,n) = A014480(n-1).

A053088 a(n) = 3*a(n-2) + 2*a(n-3) for n > 2, a(0)=1, a(1)=0, a(2)=3.

Original entry on oeis.org

1, 0, 3, 2, 9, 12, 31, 54, 117, 224, 459, 906, 1825, 3636, 7287, 14558, 29133, 58248, 116515, 233010, 466041, 932060, 1864143, 3728262, 7456549, 14913072, 29826171, 59652314, 119304657, 238609284, 477218599, 954437166, 1908874365
Offset: 0

Author

Pauline Gorman (pauline(AT)gorman65.freeserve.co.uk), Feb 26 2000

Keywords

Comments

Growth of happy bug population in GCSE math course work assignment.
The generalized (3,2)-Padovan sequence p(3,2;n). See the W. Lang link under A000931. - Wolfdieter Lang, Jun 25 2010
With offset 1: a(n) = -2^n*Sum_{k=0..n} k^p*q^k for p=1, q=-1/2. See also A232603 (p=2, q=-1/2), A232604 (p=3, q=-1/2). - Stanislav Sykora, Nov 27 2013
From Paul Curtz, Nov 02 2021 (Start)
a(n-2) difference table (from 0, 0, a(n)):
0 0 1 0 3 2 9 12 31 54 ...
0 1 -1 3 -1 7 3 19 23 63 ...
1 -2 4 -4 8 -4 16 4 40 44 ...
-3 6 -8 12 -12 20 -12 36 4 84 ...
9 -14 20 -24 32 -32 48 -32 80 0 ...
-23 34 -44 56 -64 80 -80 112 -80 176 ...
57 -78 100 -120 144 -160 192 -192 256 -192 ...
... .
The signature is valid for every row.
a(n-2) + a(n-1) = A001045(n).
a(n-2) + a(n+1) = A062510(n) = 3*A001045(n).
a(n-2) + a(n+3) = see A144472(n+1).
Second subdiagonal: 1, 6, 20, 56, 144, 352, ... = A014480(n).
First subdiagonal: -A036895(n) = -2*A001787(n).
Main diagonal: A001787(n) = -first and -third upper diagonals.
Second, fourth and fifth upper diagonals: A001792(n), A045891(n+2) and A172160(n+1). (End)

Programs

  • Mathematica
    CoefficientList[Series[1/(1 - 3 x^2 - 2 x^3), {x, 0, 32}], x] (* Michael De Vlieger, Sep 30 2019 *)
  • PARI
    c(n)=(2^(n+1)-(-1)^n*(3*n+2))/9; a(n)=c(n+1); \\ Stanislav Sykora, Nov 27 2013

Formula

G.f.: 1 / (1-3*x^2-2*x^3).
With offset 1: a(1)=1; a(n) = 2*a(n-1) - (-1)^n*n; a(n) = (1/9)*(2^(n+1) - (-1)^n*(3*n+2)). - Benoit Cloitre, Nov 02 2002
a(n) = Sum_{k=0..floor(n/2)} A078008(n-2k). - Paul Barry, Nov 24 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(k, n-2k)*3^k*(2/3)^(n-2k). - Paul Barry, Oct 16 2004
a(n) = Sum_{k=0..n} A078008(k)*(1 - (-1)^(n+k-1))/2. - Paul Barry, Apr 16 2005
a(n) = ( 2^(n+2) + (-1)^n*(3*n+5) )/9 (see also the B. Cloitre comment above). From the o.g.f. 1/(1-3*x^2-2*x^3) = 1/((1-2*x)*(1+x)^2) = (3/(1+x)^2 + 2/(1+x) + 4/(1-2*x))/9. - Wolfdieter Lang, Jun 25 2010
From Wolfdieter Lang, Aug 26 2010: (Start)
a(n) = a(n-1) + 2*a(n-2) + (-1)^n for n > 1, a(0)=1, a(1)=0.
Due to the identity for the o.g.f. A(x): A(x) = x*(1+2*x)*A(x) + 1/(1+x).
(This recurrence was observed by Gary Detlefs in a 08/25/10 e-mail to the author.) (End)
G.f.: Sum_{n>=0} binomial(3*n,n)*x^n / (1+x)^(3*n+3). - Paul D. Hanna, Mar 03 2012
E.g.f.: 1 + (1/9)*(exp(-x)*(3*x - 2) + 2*exp(2*x)). - Stefano Spezia, Sep 27 2019

Extensions

More terms from James Sellers, Feb 28 2000 and Christian G. Bower, Feb 29 2000

A073345 Table T(n,k), read by ascending antidiagonals, giving the number of rooted plane binary trees of size n and height k.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 6, 8, 0, 0, 0, 0, 0, 0, 0, 4, 20, 0, 0, 0, 0, 0, 0, 0, 0, 1, 40, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 68, 56, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 94, 152, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 114, 376, 144, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Author

Antti Karttunen, Jul 31 2002

Keywords

Examples

			The top-left corner of this square array is
  1 0 0 0 0 0 0 0 0 ...
  0 1 0 0 0 0 0 0 0 ...
  0 0 2 1 0 0 0 0 0 ...
  0 0 0 4 6 6 4 1 0 ...
  0 0 0 0 8 20 40 68 94 ...
E.g. we have A000108(3) = 5 binary trees built from 3 non-leaf (i.e. branching) nodes:
_______________________________3
___\/__\/____\/__\/____________2
__\/____\/__\/____\/____\/_\/__1
_\/____\/____\/____\/____\./___0
The first four have height 3 and the last one has height 2, thus T(3,3) = 4, T(3,2) = 1 and T(3,any other value of k) = 0.
		

References

  • Luo Jian-Jin, Catalan numbers in the history of mathematics in China, in Combinatorics and Graph Theory, (Yap, Ku, Lloyd, Wang, Editors), World Scientific, River Edge, NJ, 1995.

Crossrefs

Variant: A073346. Column sums: A000108. Row sums: A001699.
Diagonals: A073345(n, n) = A011782(n), A073345(n+3, n+2) = A014480(n), A073345(n+2, n) = A073773(n), A073345(n+3, n) = A073774(n) - Henry Bottomley and AK, see the attached notes.
A073429 gives the upper triangular region of this array. Cf. also A065329, A001263.

Programs

  • Maple
    A073345 := n -> A073345bi(A025581(n), A002262(n));
    A073345bi := proc(n,k) option remember; local i,j; if(0 = n) then if(0 = k) then RETURN(1); else RETURN(0); fi; fi; if(0 = k) then RETURN(0); fi; 2 * add(A073345bi(n-i-1,k-1) * add(A073345bi(i,j),j=0..(k-1)),i=0..floor((n-1)/2)) + 2 * add(A073345bi(n-i-1,k-1) * add(A073345bi(i,j),j=0..(k-2)),i=(floor((n-1)/2)+1)..(n-1)) - (`mod`(n,2))*(A073345bi(floor((n-1)/2),k-1)^2); end;
    A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1);
    A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);
  • Mathematica
    a[0, 0] = 1; a[n_, k_]/;k2^n-1 := 0; a[n_, k_]/;1 <= n <= k <= 2^n-1 := a[n, k] = Sum[a[n-1, k-1-i](2Sum[ a[j, i], {j, 0, n-2}]+a[n-1, i]), {i, 0, k-1}]; Table[a[n, k], {n, 0, 9}, {k, 0, 9}]
    (* or *) a[0] = 0; a[1] = 1; a[n_]/;n>=2 := a[n] = Expand[1 + x a[n-1]^2]; gfT[n_] := a[n]-a[n-1]; Map[CoefficientList[ #, x, 8]&, Table[gfT[n], {n, 9}]/.{x^i_/;i>=9 ->0}] (Callan)

Formula

(See the Maple code below. Is there a nicer formula?)
This table was known to the Chinese mathematician Ming An-Tu, who gave the following recurrence in the 1730s. a(0, 0) = 1, a(n, k) = Sum[a(n-1, k-1-i)( 2*Sum[ a(j, i), {j, 0, n-2}]+a(n-1, i) ), {i, 0, k-1}]. - David Callan, Aug 17 2004
The generating function for row n, T_n(x):=Sum[T(n, k)x^k, k>=0], is given by T_n = a(n)-a(n-1) where a(n) is defined by the recurrence a(0)=0, a(1)=1, a(n) = 1 + x a(n-1)^2 for n>=2. - David Callan, Oct 08 2005
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