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.

Showing 1-9 of 9 results.

A285103 Number of odd terms on row n of A053632: a(n) = A000120(A068052(n)).

Original entry on oeis.org

1, 2, 4, 6, 6, 12, 12, 16, 22, 28, 32, 30, 36, 52, 48, 62, 62, 68, 88, 104, 116, 108, 128, 128, 132, 168, 160, 168, 200, 204, 240, 232, 242, 284, 300, 324, 332, 348, 352, 352, 412, 440, 400, 466, 460, 516, 496, 566, 582, 580, 608, 646, 676, 736, 716, 782, 728, 816, 832, 856, 916, 924, 948, 1034, 1008, 1044, 1096, 1154, 1112, 1212, 1204, 1188
Offset: 0

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Author

Antti Karttunen, Apr 15 2017

Keywords

Crossrefs

Number of odd term on row n of A053632.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
          (t-> Bits[Xor](2^n*t, t))(b(n-1)))
        end:
    a:= n-> convert(Bits[Split](b(n)), `+`):
    seq(a(n), n=0..71);  # Alois P. Heinz, Mar 07 2024
  • Mathematica
    b[n_] := b[n] = If[n == 0, 1, With[{t = b[n-1]}, BitXor[2^n*t, t]]];
    a[n_] := DigitCount[b[n], 2, 1];
    Table[a[n], {n, 0, 100}] (* Jean-François Alcover, May 17 2024, after Alois P. Heinz *)
  • Python
    # uses [A000120]
    l=[1]
    for n in range(1, 101):
        x = l[n - 1]
        l.append(x^(2**n*x))
    print([A000120(k) for k in l]) # Indranil Ghosh, Jun 28 2017
  • Scheme
    (define (A285103 n) (A000120 (A068052 n)))
    

Formula

a(n) = A000120(A068052(n)).
a(n) = A001221(A285102(n)) = A001222(A285102(n)).
A285104(n) = 2^n - a(n).
A000124(n) = a(n) + A285105(n).

A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

Original entry on oeis.org

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012
From R. K. Guy, Dec 28 2005: (Start)
"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):
0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...
0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...
.....-.-.....+..+.....-..-.....+..+......-...-.......+....
"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.
"Append signs according as the pair have the same (+) or opposite (-) parity.
"Then Euler's pentagonal number theorem is easy to remember:
"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n
where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.
"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."
(End)
The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003
Number of levels in the partitions of n + 1 with parts in {1,2}.
a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007
Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008
From Matthew Vandermast, Oct 28 2008: (Start)
Numbers n for which A000326(n) is a member of A000332. Cf. A145920.
This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)
Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009
Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010
Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011
A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012
a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013
Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013
Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014
(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014
Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016
The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018
Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018
Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018
The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018
Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022
8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022
a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022
Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023
From Peter Bala, Jan 06 2025: (Start)
The sequence terms are the exponents in the expansions of the following infinite products:
1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).
2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....
3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....
4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...
		

References

  • Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
  • 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

Cf. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).
Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).
Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.
Union of A036498 and A036499.
Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).
Column 1 of A195152.
Squares in APs: A221671, A221672.
Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.
Cf. A002620.

Programs

  • GAP
    a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018
    
  • Haskell
    a001318 n = a001318_list !! n
    a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015
    
  • Magma
    [(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014
    
  • Magma
    [(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014
    
  • Maple
    A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
    A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011
  • Mathematica
    Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
    Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
    CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
    a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)
  • PARI
    {a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */
    
  • Python
    def a(n):
        p = n % 2
        return (n + p)*(3*n + 2 - p) >> 3
    print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022
    
  • Python
    def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024
  • Sage
    @CachedFunction
    def A001318(n):
        if n == 0 : return 0
        inc = n//2 if is_even(n) else n
        return inc + A001318(n-1)
    [A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012
    

Formula

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).
A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006
Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011
a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]
a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011
Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011
G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams
a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003
Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003
a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005
a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005
a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007
If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007
a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011
a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012
a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013
a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A026741(n),
a(n) = a(n+2) - A001651(n),
a(n) = a(n+3) - A184418(n),
a(n) = a(n+4) - A007310(n),
a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),
a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),
a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),
a(n) = a(n+12)- A007310(n)*3,
a(n) = a(n+14)- A001651(n)*7. (End)
a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018
a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013
G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016
a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018
a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018
a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019
Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021
a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022
a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024

A048675 If n = p_i^e_i * ... * p_k^e_k, p_i < ... < p_k primes (with p_i = prime(i)), then a(n) = (1/2) * (e_i * 2^i + ... + e_k * 2^k).

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 8, 3, 4, 5, 16, 4, 32, 9, 6, 4, 64, 5, 128, 6, 10, 17, 256, 5, 8, 33, 6, 10, 512, 7, 1024, 5, 18, 65, 12, 6, 2048, 129, 34, 7, 4096, 11, 8192, 18, 8, 257, 16384, 6, 16, 9, 66, 34, 32768, 7, 20, 11, 130, 513, 65536, 8, 131072, 1025, 12, 6, 36, 19
Offset: 1

Views

Author

Antti Karttunen, Jul 14 1999

Keywords

Comments

The original motivation for this sequence was to encode the prime factorization of n in the binary representation of a(n), each such representation being unique as long as this map is restricted to A005117 (squarefree numbers, resulting a permutation of nonnegative integers A048672) or any of its subsequence, resulting an injective function like A048623 and A048639.
However, also the restriction to A260443 (not all terms of which are squarefree) results a permutation of nonnegative integers, namely A001477, the identity permutation.
When a polynomial with nonnegative integer coefficients is encoded with the prime factorization of n (e.g., as in A206296, A260443), then a(n) gives the evaluation of that polynomial at x=2.
The primitive completely additive integer sequence that satisfies a(n) = a(A225546(n)), n >= 1. By primitive, we mean that if b is another such sequence, then there is an integer k such that b(n) = k * a(n) for all n >= 1. - Peter Munn, Feb 03 2020
If the binary rank of an integer partition y is given by Sum_i 2^(y_i-1), and the Heinz number is Product_i prime(y_i), then a(n) is the binary rank of the integer partition with Heinz number n. Note the function taking a set s to Sum_i 2^(s_i-1) is the inverse of A048793 (binary indices), and the function taking a multiset m to Product_i prime(m_i) is the inverse of A112798 (prime indices). - Gus Wiseman, May 22 2024

Examples

			From _Gus Wiseman_, May 22 2024: (Start)
The A018819(7) = 6 cases of binary rank 7 are the following, together with their prime indices:
   30: {1,2,3}
   40: {1,1,1,3}
   54: {1,2,2,2}
   72: {1,1,1,2,2}
   96: {1,1,1,1,1,2}
  128: {1,1,1,1,1,1,1}
(End)
		

Crossrefs

Row 2 of A104244.
Similar logarithmic functions: A001414, A056239, A090880, A289506, A293447.
Left inverse of the following sequences: A000079, A019565, A038754, A068911, A134683, A260443, A332824.
A003961, A028234, A032742, A055396, A064989, A067029, A225546, A297845 are used to express relationship between terms of this sequence.
Cf. also A048623, A048676, A099884, A277896 and tables A277905, A285325.
Cf. A297108 (Möbius transform), A332813 and A332823 [= a(n) mod 3].
Pairs of sequences (f,g) that satisfy a(f(n)) = g(n), possibly with offset change: (A000203,A331750), (A005940,A087808), (A007913,A248663), (A007947,A087207), (A097248,A048675), (A206296,A000129), (A248692,A056239), (A283477,A005187), (A284003,A006068), (A285101,A028362), (A285102,A068052), (A293214,A001065), (A318834,A051953), (A319991,A293897), (A319992,A293898), (A320017,A318674), (A329352,A069359), (A332461,A156552), (A332462,A156552), (A332825,A000010) and apparently (A163511,A135529).
See comments/formulas in A277333, A331591, A331740 giving their relationship to this sequence.
The formula section details how the sequence maps the terms of A329050, A329332.
A277892, A322812, A322869, A324573, A324575 give properties of the n-th term of this sequence.
The term k appears A018819(k) times.
The inverse transformation is A019565 (Heinz number of binary indices).
The version for distinct prime indices is A087207.
Numbers k such that a(k) is prime are A277319, counts A372688.
Grouping by image gives A277905.
A014499 lists binary indices of prime numbers.
A061395 gives greatest prime index, least A055396.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.
Binary indices:
- listed A048793, sum A029931
- reversed A272020
- opposite A371572, sum A230877
- length A000120, complement A023416
- min A001511, opposite A000012
- max A070939, opposite A070940
- complement A368494, sum A359400
- opposite complement A371571, sum A359359

Programs

  • Maple
    nthprime := proc(n) local i; if(isprime(n)) then for i from 1 to 1000000 do if(ithprime(i) = n) then RETURN(i); fi; od; else RETURN(0); fi; end; # nthprime(2) = 1, nthprime(3) = 2, nthprime(5) = 3, etc. - this is also A049084.
    A048675 := proc(n) local s,d; s := 0; for d in ifactors(n)[ 2 ] do s := s + d[ 2 ]*(2^(nthprime(d[ 1 ])-1)); od; RETURN(s); end;
    # simpler alternative
    f:= n -> add(2^(numtheory:-pi(t[1])-1)*t[2], t=ifactors(n)[2]):
    map(f, [$1..100]); # Robert Israel, Oct 10 2016
  • Mathematica
    a[1] = 0; a[n_] := Total[ #[[2]]*2^(PrimePi[#[[1]]]-1)& /@ FactorInteger[n] ]; Array[a, 100] (* Jean-François Alcover, Mar 15 2016 *)
  • PARI
    a(n) = my(f = factor(n)); sum(k=1, #f~, f[k,2]*2^primepi(f[k,1]))/2; \\ Michel Marcus, Oct 10 2016
    
  • PARI
    \\ The following program reconstructs terms (e.g. for checking purposes) from the factorization file prepared by Hans Havermann:
    v048675sigs = readvec("a048675.txt");
    A048675(n) = if(n<=2,n-1,my(prsig=v048675sigs[n],ps=prsig[1],es=prsig[2]); prod(i=1,#ps,ps[i]^es[i])); \\ Antti Karttunen, Feb 02 2020
    
  • Python
    from sympy import factorint, primepi
    def a(n):
        if n==1: return 0
        f=factorint(n)
        return sum([f[i]*2**(primepi(i) - 1) for i in f])
    print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jun 19 2017

Formula

a(1) = 0, a(n) = 1/2 * (e1*2^i1 + e2*2^i2 + ... + ez*2^iz) if n = p_{i1}^e1*p_{i2}^e2*...*p_{iz}^ez, where p_i is the i-th prime. (e.g. p_1 = 2, p_2 = 3).
Totally additive with a(p^e) = e * 2^(PrimePi(p)-1), where PrimePi(n) = A000720(n). [Missing factor e added to the comment by Antti Karttunen, Jul 29 2015]
From Antti Karttunen, Jul 29 2015: (Start)
a(1) = 0; for n > 1, a(n) = 2^(A055396(n)-1) + a(A032742(n)). [Where A055396(n) gives the index of the smallest prime dividing n and A032742(n) gives the largest proper divisor of n.]
a(1) = 0; for n > 1, a(n) = (A067029(n) * (2^(A055396(n)-1))) + a(A028234(n)).
Other identities. For all n >= 0:
a(A019565(n)) = n.
a(A260443(n)) = n.
a(A206296(n)) = A000129(n).
a(A005940(n+1)) = A087808(n).
a(A007913(n)) = A248663(n).
a(A007947(n)) = A087207(n).
a(A283477(n)) = A005187(n).
a(A284003(n)) = A006068(n).
a(A285101(n)) = A028362(1+n).
a(A285102(n)) = A068052(n).
Also, it seems that a(A163511(n)) = A135529(n) for n >= 1. (End)
a(1) = 0, a(2n) = 1+a(n), a(2n+1) = 2*a(A064989(2n+1)). - Antti Karttunen, Oct 11 2016
From Peter Munn, Jan 31 2020: (Start)
a(n^2) = a(A003961(n)) = 2 * a(n).
a(A297845(n,k)) = a(n) * a(k).
a(n) = a(A225546(n)).
a(A329332(n,k)) = n * k.
a(A329050(n,k)) = 2^(n+k).
(End)
From Antti Karttunen, Feb 02-25 2020, Feb 01 2021: (Start)
a(n) = Sum_{d|n} A297108(d) = Sum_{d|A225546(n)} A297108(d).
a(n) = a(A097248(n)).
For n >= 2:
A001221(a(n)) = A322812(n), A001222(a(n)) = A277892(n).
A000203(a(n)) = A324573(n), A033879(a(n)) = A324575(n).
For n >= 1, A331750(n) = a(A000203(n)).
For n >= 1, the following chains hold:
A293447(n) >= a(n) >= A331740(n) >= A331591(n).
a(n) >= A087207(n) >= A248663(n).
(End)
a(n) = A087207(A097248(n)). - Flávio V. Fernandes, Jul 16 2025

Extensions

Entry revised by Antti Karttunen, Jul 29 2015
More linking formulas added by Antti Karttunen, Apr 18 2017

A053632 Irregular triangle read by rows giving coefficients in expansion of Product_{k=1..n} (1 + x^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 7, 7, 6, 5, 5, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 4
Offset: 0

Views

Author

N. J. A. Sloane, Mar 22 2000

Keywords

Comments

Or, triangle T(n,k) read by rows, giving number of subsets of {1,2,...,n} with sum k. - Roger CUCULIERE (cuculier(AT)imaginet.fr), Nov 19 2000
Row n consists of A000124(n) terms. These are also the successive vectors (their nonzero elements) when one starts with the infinite vector (of zeros) with 1 inserted somewhere and then shifts it one step (right or left) and adds to the original, then shifts the result two steps and adds, three steps and adds, etc. - Antti Karttunen, Feb 13 2002
T(n,k) = number of partitions of k into distinct parts <= n. Triangle of distribution of Wilcoxon's signed rank statistic. - Mitch Harris, Mar 23 2006
T(n,k) = number of binary words of length n in which the sum of the positions of the 0's is k. Example: T(4,5)=2 because we have 0110 (sum of the positions of the 0's is 1+4=5) and 1001 (sum of the positions of the 0's is 2+3=5). - Emeric Deutsch, Jul 23 2006
A fair coin is flipped n times. You receive i dollars for a "success" on the i-th flip, 1<=i<=n. T(n,k)/2^n is the probability that you will receive exactly k dollars. Your expectation is n(n+1)/4 dollars. - Geoffrey Critzer, May 16 2010
From Gus Wiseman, Jan 02 2023: (Start)
With offset 1, also the number of integer compositions of n whose partial sums add up to k for k = n..n(n+1)/2. For example, row n = 6 counts the following compositions:
6 15 24 33 42 51 141 231 321 411 1311 2211 3111 12111 21111 111111
114 123 132 222 312 1131 1221 2121 11121 11211
213 1113 1122 1212 2112 1111
(End)

Examples

			Triangle begins:
  1;
  1, 1;
  1, 1, 1, 1;
  1, 1, 1, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 1, 1, 1;
  1, 1, 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 3, 2, 2, 1, 1, 1;
  ...
Row n = 4 counts the following binary words, where k = sum of positions of zeros:
  1111  0111  1011  0011  0101  0110  0001  0010  0100  1000  0000
                    1101  1110  1001  1010  1100
Row n = 5 counts the following strict partitions of k with all parts <= n (0 is the empty partition):
  0  1  2  3  4  5  42  43  53  54  532  542  543  5431 5432 54321
           21 31 32 51  52  431 432 541  5321 5421
                 41 321 421 521 531 4321
		

References

  • A. V. Yurkin, New binomial and new view on light theory, (book), 2013, 78 pages, no publisher listed.

Crossrefs

Rows reduced modulo 2 and interpreted as binary numbers: A068052, A068053. Rows converge towards A000009.
Row sums give A000079.
Cf. A285101 (multiplicative encoding of each row), A285103 (number of odd terms on row n), A285105 (number of even terms).
Row lengths are A000124.
A reciprocal version is (A033999, A219977, A291983, A291984, A291985, ...).
A negative version is A231599.
A version for partitions is A358194, reversed partitions A264034.

Programs

  • Maple
    with(gfun,seriestolist); map(op,[seq(seriestolist(series(mul(1+(z^i), i=1..n),z,binomial(n+1,2)+1)), n=0..10)]); # Antti Karttunen, Feb 13 2002
    # second Maple program:
    g:= proc(n) g(n):= `if`(n=0, 1, expand(g(n-1)*(1+x^n))) end:
    T:= n-> seq(coeff(g(n), x, k), k=0..degree(g(n))):
    seq(T(n), n=0..10);  # Alois P. Heinz, Nov 19 2012
  • Mathematica
    Table[CoefficientList[ Series[Product[(1 + t^i), {i, 1, n}], {t, 0, 100}], t], {n, 0, 8}] // Grid (* Geoffrey Critzer, May 16 2010 *)

Formula

From Mitch Harris, Mar 23 2006: (Start)
T(n,k) = T(n-1, k) + T(n-1, k-n), T(0,0)=1, T(0,k) = 0, T(n,k) = 0 if k < 0 or k > (n+1 choose 2).
G.f.: (1+x)*(1+x^2)*...*(1+x^n). (End)
Sum_{k>=0} k * T(n,k) = A001788(n). - Alois P. Heinz, Feb 09 2017
max_{k>=0} T(n,k) = A025591(n). - Alois P. Heinz, Jan 20 2023

A028362 Total number of self-dual binary codes of length 2n. Totally isotropic spaces of index n in symplectic geometry of dimension 2n.

Original entry on oeis.org

1, 3, 15, 135, 2295, 75735, 4922775, 635037975, 163204759575, 83724041661975, 85817142703524375, 175839325399521444375, 720413716161839357604375, 5902349576513949856852644375, 96709997811181068404530578084375
Offset: 1

Views

Author

Keywords

Comments

These numbers appear in the second column of A155103. - Mats Granvik, Jan 20 2009
a(n) = n terms in the sequence (1, 2, 4, 8, 16, ...) dot n terms in the sequence (1, 1, 3, 15, 135). Example: a(5) = 2295 = (1, 2, 4, 8, 16) dot (1, 1, 3, 15, 135) = (1 + 2 + 12 + 120 + 2160). - Gary W. Adamson, Aug 02 2010

Examples

			G.f. = x + 3*x^2 + 15*x^3 + 135*x^4 + 2295*x^5 + 75735*x^6 + 4922775*x^7 + ...
		

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 630.

Crossrefs

Cf. A155103. - Mats Granvik, Jan 20 2009
Cf. A005329, A006088. - Paul D. Hanna, Sep 16 2009

Programs

  • Magma
    [1] cat [&*[ 2^k+1: k in [1..n] ]: n in [1..16]]; // Vincenzo Librandi, Dec 24 2015
    
  • Maple
    seq(mul(1 + 2^j, j = 1..n-1), n = 1..20); # G. C. Greubel, Jun 06 2020
  • Mathematica
    Table[Product[2^i+1,{i,n-1}],{n,15}] (* or *) FoldList[Times,1, 2^Range[15]+1] (* Harvey P. Dale, Nov 21 2011 *)
    Table[QPochhammer[-2, 2, n - 1], {n, 15}] (* Arkadiusz Wesolowski, Oct 29 2012 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,2^(m*(m-1)/2)*x^m/prod(k=0,m-1,1-2^k*x+x*O(x^n))),n)} \\ Paul D. Hanna, Sep 16 2009
    
  • PARI
    {a(n) = if( n<1, 0 , prod(k=1, n-1, 2^k + 1))}; /* Michael Somos, Jan 28 2018 */
    
  • PARI
    {a(n) = sum(k=0, n-1, 2^(k*(k+1)/2) * prod(j=1, k, (2^(n-j) - 1) / (2^j - 1)))}; /* Michael Somos, Jan 28 2018 */
    
  • Python
    for n in range(2,40,2):
      product = 1
      for i in range(1,n//2-1 + 1):
        product *= (2**i+1)
      print(product)
    # Nathan J. Russell, Mar 01 2016
    
  • Python
    from math import prod
    def A028362(n): return prod((1<Chai Wah Wu, Jun 20 2022
    
  • Sage
    from ore_algebra import *
    R. = QQ['x']
    A. = OreAlgebra(R, 'Qx', q=2)
    print((Qx - x - 1).to_list([0,1], 10))  # Ralf Stephan, Apr 24 2014
    
  • Sage
    from sage.combinat.q_analogues import q_pochhammer
    [q_pochhammer(n-1,-2,2) for n in (1..20)] # G. C. Greubel, Jun 06 2020
    
  • Scheme
    ;; With memoization-macro definec.
    (define (A028362 n) (A028362off0 (- n 1)))
    (definec (A028362off0 n) (if (zero? n) 1 (+ (A028362off0 (- n 1)) (* (expt 2 n) (A028362off0 (- n 1))))))
    ;; Antti Karttunen, Apr 15 2017

Formula

a(n) = Product_{i=1..n-1} (2^i+1).
Letting a(0)=1, we have a(n) = Sum_{k=0..n-1} 2^k*a(k) for n>0. a(n) is asymptotic to c*sqrt(2)^(n^2-n) where c=2.384231029031371724149899288.... = A079555 = Product_{k>=1} (1 + 1/2^k). - Benoit Cloitre, Jan 25 2003
G.f.: Sum_{n>=1} 2^(n*(n-1)/2) * x^n/(Product_{k=0..n-1} (1-2^k*x)). - Paul D. Hanna, Sep 16 2009
a(n) = 2^(binomial(n,2) - 1)*(-1; 1/2){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015
From Antti Karttunen, Apr 15 2017: (Start)
a(n) = A048675(A285101(n-1)).
a(n) = b(n-1), where b(0) = 1, and for n > 0, b(n) = b(n-1) + (2^n)*b(n-1).
a(n) = Sum_{i=1..A000124(n-1)} A053632(n-1,i-1)*(2^(i-1)) [where the indexing of both rows and columns of irregular table A053632(row,col) is considered to start from zero].
(End)
G.f. A(x) satisfies: A(x) = x * (1 + A(2*x)) / (1 - x). - Ilya Gutkovskiy, Jun 06 2020
Conjectural o.g.f. as a continued fraction of Stieltjes type (S-fraction):
1/(1 - 3*x/(1 - 2*x/(1 - 10*x/(1 - 12*x/(1 - 36*x/(1 - 56*x/(1 - 136*x/(1 - 240*x/(1 - ... - 2^(n-1)*(2^n + 1)*x/(1 - 2^n*(2^n - 1)*x/(1 - ... ))))))))))). - Peter Bala, Sep 27 2023

A285101 a(0) = 2, for n > 0, a(n) = a(n-1)*A242378(n,a(n-1)), where A242378(n,a(n-1)) shifts the prime factorization of a(n-1) n primes towards larger primes with A003961.

Original entry on oeis.org

2, 6, 210, 3573570, 64845819350301990, 28695662573739152697846686144187168109530, 1038300112150956151877699324649731518883355380534272386781875587619359740733888844803014212990
Offset: 0

Views

Author

Antti Karttunen, Apr 15 2017

Keywords

Comments

Multiplicative encoding of irregular table A053632 (in style of A007188 and A260443).

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A242378(k,n) = { while(k>0,n = A003961(n); k = k-1); n; };
    A285101(n) = { if(0==n,2,A285101(n-1)*A242378(n,A285101(n-1))); };
    
  • Python
    from sympy import factorint, prime, primepi
    from operator import mul
    from functools import reduce
    def a003961(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**f[i] for i in f])
    def a242378(k, n):
        while k>0:
            n=a003961(n)
            k-=1
        return n
    l=[2]
    for n in range(1, 7):
        x=l[n - 1]
        l.append(x*a242378(n, x))
    print(l) # Indranil Ghosh, Jun 27 2017
  • Scheme
    (definec (A285101 n) (if (zero? n) 2 (* (A285101 (- n 1)) (A242378bi n (A285101 (- n 1)))))) ;; For A242378bi see A242378.
    

Formula

a(0) = 2, for n > 0, a(n) = a(n-1)*A242378(n,a(n-1)).
Other identities. For all n >= 0:
A001222(a(n)) = A000079(n).
A048675(a(n)) = A028362(1+n).
A248663(a(n)) = A068052(n).
A007913(a(n)) = A285102(n).

A285102 a(n) = A007913(A285101(n)).

Original entry on oeis.org

2, 6, 210, 72930, 620310, 278995269860970, 12849025509071310, 492608110538467706074890, 1342951001046021018427857601026746070, 37793589449865555275592120894959094883390892772270, 728982633030274864467458719371654181886452163442582606072870, 28339554655955912942523491885490197708224606885407444005070
Offset: 0

Views

Author

Antti Karttunen, Apr 15 2017

Keywords

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A242378(k,n) = { while(k>0,n = A003961(n); k = k-1); n; };
    A285102(n) = { if(0==n,2,lcm(A285102(n-1),A242378(n,A285102(n-1)))/gcd(A285102(n-1),A242378(n,A285102(n-1)))); };
    
  • Python
    # uses [A003961, A242378]
    from sympy import factorint, prime, primepi
    from sympy.ntheory.factor_ import core
    from operator import mul
    def a003961(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**f[i] for i in f])
    def a242378(k, n):
        while k>0:
            n=a003961(n)
            k-=1
        return n
    l=[2]
    for n in range(1, 12):
        x=l[n - 1]
        l.append(x*a242378(n, x))
    print([core(j) for j in l]) # Indranil Ghosh, Jun 27 2017
  • Scheme
    (definec (A285102 n) (if (zero? n) 2 (/ (lcm (A285102 (- n 1)) (A242378bi n (A285102 (- n 1)))) (gcd (A285102 (- n 1)) (A242378bi n (A285102 (- n 1)))))))
    

Formula

a(0) = 2, for n > 0, a(n) = lcm(a(n-1),A242378(n,a(n-1))) / gcd(a(n-1),A242378(n,a(n-1))).
a(n) = A007913(A285101(n)).
Other identities. For all n >= 0:
A001221(a(n)) = A001222(a(n)) = A285103(n).
A048675(a(n)) = A068052(n).

A285105 Number of even terms on row n of A053632.

Original entry on oeis.org

0, 0, 0, 1, 5, 4, 10, 13, 15, 18, 24, 37, 43, 40, 58, 59, 75, 86, 84, 87, 95, 124, 126, 149, 169, 158, 192, 211, 207, 232, 226, 265, 287, 278, 296, 307, 335, 356, 390, 429, 409, 422, 504, 481, 531, 520, 586, 563, 595, 646, 668, 681, 703, 696, 770, 759, 869, 838, 880, 915, 915, 968, 1006, 983, 1073, 1102, 1116, 1125, 1235, 1204, 1282, 1369, 1349
Offset: 0

Views

Author

Antti Karttunen, Apr 15 2017

Keywords

Crossrefs

Programs

Formula

a(n) = A080791(A068052(n)).
A000124(n) = A285103(n) + a(n).

A068053 Start with one 1 in the infinite vector of zeros, shift one right or left and sum mod 2 (bitwise-XOR) to get 11, then shift two steps and XOR to get 1111, then three steps and XOR to get 1110111, then four steps and so on.

Original entry on oeis.org

1, 11, 1111, 1110111, 11100000111, 1110011111100111, 1110010001111000100111, 11100101101100000110110100111, 1110010101010101110111010101010100111, 1110010100100111011101111011101110010010100111
Offset: 0

Views

Author

Antti Karttunen, Feb 13 2002

Keywords

Crossrefs

Sequence A068052 converted to binary. Rows of A053632 reduced mod 2 and concatenated.

Programs

  • Maple
    map(convert, A068052,binary);
  • Mathematica
    Map[FromDigits[IntegerDigits[#, 2]] &, FoldList[BitXor[#, #*#2] &, 1, 2^Range[10]]] (* Paolo Xausa, Mar 07 2024 *)
Showing 1-9 of 9 results.