A005891 -id:A005891 - cp's OEIS frontend

A280952 Expansion of Product_{k>=0} 1/(1 - x^(5k(k+1)/2+1)).

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 10, 11, 11, 12, 12, 13, 14, 15, 15, 16, 16, 17, 18, 19, 19, 20, 21, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 37, 38, 40, 42, 44, 45, 47, 49, 51, 53, 55, 56, 58, 60, 62, 64, 67, 68, 71, 74, 77, 79, 83, 85, 88, 91, 94, 96, 100

Offset: 0

Ilya Gutkovskiy, Jan 11 2017

nonn

Number of partitions of n into centered pentagonal numbers (A005891).

			a(12) = 3 because we have [6, 6], [6, 1, 1, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1].

Alois P. Heinz, Table of n, a(n) for n = 0..20000
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to arXiv version]
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for sequences related to centered polygonal numbers
Index entries for related partition-counting sequences

Cf. A005891, A218379, A279221, A280950, A280951, A280953.

h:= proc(n) option remember; `if`(n<0, 0, (t->
      `if`(((t+1)*5*t+2)/2>n, t-1, t))(1+h(n-1)))
    end:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      b(n, i-1)+(t-> b(n-t, min(i, h(n-t))))(((i+1)*5*i+2)/2)))
    end:
a:= n-> b(n, h(n)):
seq(a(n), n=0..100);  # Alois P. Heinz, Dec 28 2018

nmax = 88; CoefficientList[Series[Product[1/(1 - x^(5 k (k + 1)/2 + 1)), {k, 0, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=0} 1/(1 - x^(5*k*(k+1)/2+1)).

A069128 Centered 15-gonal numbers: a(n) = (15n^2 - 15n + 2)/2.

Original entry on oeis.org

1, 16, 46, 91, 151, 226, 316, 421, 541, 676, 826, 991, 1171, 1366, 1576, 1801, 2041, 2296, 2566, 2851, 3151, 3466, 3796, 4141, 4501, 4876, 5266, 5671, 6091, 6526, 6976, 7441, 7921, 8416, 8926, 9451, 9991, 10546, 11116, 11701, 12301, 12916, 13546, 14191, 14851, 15526

Offset: 1

Terrel Trotter, Jr., Apr 07 2002

Centered pentadecagonal numbers or centered quindecagonal numbers or centered pentakaidecagonal numbers. - Omar E. Pol, Oct 03 2011

			a(5) = 151 because (15*5^2 - 15*5 + 2)/2 = 151.

T. D. Noe, Table of n, a(n) for n = 1..1000
E. Weisstein, Centered Polygonal Numbers
Index entries for sequences related to centered polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005448, A001844, A005891, A003215, A069099.

[(15*n^2 - 15*n + 2)/2 : n in [1..50]]; // Wesley Ivan Hurt, Nov 14 2014

A069128:=n->(15*n^2 - 15*n + 2)/2: seq(A069128(n), n=1..50); # Wesley Ivan Hurt, Nov 14 2014

FoldList[#1 + #2 &, 1, 15 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
LinearRecurrence[{3,-3,1},{1,16,46},50] (* Harvey P. Dale, Oct 22 2013 *)

a(n)=15*n*(n-1)/2+1 \\ Charles R Greathouse IV, Nov 15 2014

a(n) = (15*n^2 - 15*n + 2)/2.
a(n) = 15*n+a(n-1)-15 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+13*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
Binomial transform of [1, 15, 15, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 15, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
a(n) = A194715(n-1) + 1. - Omar E. Pol, Oct 03 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(7/15)*Pi/2)/sqrt(105).
Sum_{n>=1} a(n)/n! = 17*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 17/(2*e) - 1. (End)
E.g.f.: exp(x)*(1 + 15*x^2/2) - 1. - Nikolaos Pantelidis, Feb 07 2023

A212656 a(n) = 5*n^2 + 1.

Original entry on oeis.org

1, 6, 21, 46, 81, 126, 181, 246, 321, 406, 501, 606, 721, 846, 981, 1126, 1281, 1446, 1621, 1806, 2001, 2206, 2421, 2646, 2881, 3126, 3381, 3646, 3921, 4206, 4501, 4806, 5121, 5446, 5781, 6126, 6481, 6846, 7221, 7606, 8001, 8406, 8821, 9246, 9681, 10126, 10581, 11046, 11521, 12006, 12501

Offset: 0

Alonso del Arte, May 23 2012

Z[sqrt(-5)] is not a unique factorization domain, and some of the numbers in this sequence have two different factorizations in that domain, e.g., 21 = 3 * 7 = (1 + 2*sqrt(-5))*(1 - 2*sqrt(-5)). And of course some primes in Z are composite in Z[sqrt(-5)], like 181 = (1 + 6*sqrt(-5))*(1 - 6*sqrt(-5)).

These are pentagonal-star numbers. - Mario Cortés, Oct 26 2020

Benjamin Fine & Gerhard Rosenberger, Number Theory: An Introduction via the Distribution of Primes, Boston: Birkhäuser, 2007, page 268.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033429, A134538, A002522, A053755, A056107, A058331.

Cf. A137530 (primes of the form 1+5*n^2).

[5*n^2 + 1: n in [0..50]]; // Vincenzo Librandi, Jul 10 2012

Table[5n^2 + 1, {n, 0, 49}]
LinearRecurrence[{3,-3,1},{1,6,21},60] (* Harvey P. Dale, Apr 04 2017 *)

a(n)=5*n^2+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 5*n^2 + 1 = (1 + n*sqrt(-5))*(1 - n*sqrt(-5)).
G.f.: (1+3*x+6*x^2)/(1-x)^3. - Bruno Berselli, May 23 2012
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Vincenzo Librandi, Jul 10 2012
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(5))*coth(Pi/sqrt(5)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(5))*csch(Pi/sqrt(5)))/2. (End)
a(n) = A005891(n-1) + 5*A000217(n). - Mario Cortés, Oct 26 2020
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(5))*sinh(sqrt(2/5)*Pi).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(5))*csch(Pi/sqrt(5)).(End)
E.g.f.: exp(x)*(1 + 5*x + 5*x^2). - Stefano Spezia, Feb 05 2021

A281083 Expansion of Product_{k>=0} (1 + x^(5k(k+1)/2+1)).

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 1

Offset: 0

Ilya Gutkovskiy, Jan 14 2017

nonn

Number of partitions of n into distinct centered pentagonal numbers (A005891).

			a(82) = 2 because we have [76, 6] and [51, 31].

Alois P. Heinz, Table of n, a(n) for n = 0..20000 (first 1001 terms from G. C. Greubel)
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for sequences related to centered polygonal numbers
Index entries for related partition-counting sequences

Cf. A005891, A218380, A280952, A281081, A281082, A281084.

nmax = 105; CoefficientList[Series[Product[1 + x^(5 k (k + 1)/2 + 1), {k, 0, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=0} (1 + x^(5*k*(k+1)/2+1)).

A348210 Varma's Kosta numbers of semi-standard tableaux: array A(n>=2, k>=0) read by rising antidiagonals.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 15, 16, 7, 1, 0, 1, 36, 65, 31, 9, 1, 0, 1, 91, 260, 175, 51, 11, 1, 0, 1, 232, 1085, 981, 369, 76, 13, 1, 0, 1, 603, 4600, 5719, 2661, 671, 106, 15, 1, 0, 1, 1585, 19845, 33922, 19929, 5916, 1105, 141, 17, 1, 0, 1, 4213, 86725, 204687, 151936, 54131, 11516, 1695, 181, 19, 1, 0

Offset: 2

R. J. Mathar, Oct 07 2021

(More characteristic NAME desired.)

Each row is a polynomial in k, which implies that the inverse binomial transformation of each row is a finite sequence and that the row can be represented by a rational generating function (A348211).

			The array starts in row n=2 with columns k>=0 as:
  0   0    0    0     0     0      0      0 ...
  1   1    1    1     1     1      1      1 ...
  1   3    5    7     9    11     13     15 ...
  1   6   16   31    51    76    106    141 ...
  1  15   65  175   369   671   1105   1695 ...
  1  36  260  981  2661  5916  11516  20385 ...
  1  91 1085 5719 19929 54131 124501 254255 ...
Antidiagonal rows begin as:
  0;
  1,   0;
  1,   1,    0;
  1,   3,    1,    0;
  1,   6,    5,    1,    0;
  1,  15,   16,    7,    1,    0;
  1,  36,   65,   31,    9,    1,   0;
  1,  91,  260,  175,   51,   11,   1,   0;
  1, 232, 1085,  981,  369,   76,  13,   1,  0;
  1, 603, 4600, 5719, 2661,  671, 106,  15,  1,  0;

G. C. Greubel, Antidiagonals n = 2..52, flattened
D.-N. Verma, Towards Classifying Finite Point-Set Configurations, 1997, Unpublished. [Scanned copy of annotated version of preprint given to me by the author in 1997. - _N. J. A. Sloane_, Oct 03 2021]

Cf. A005043 (column k=1), A007043 (k=2), A264608 (k=3), A272393 (k=4), A005408 (row n=4), A005891 (n=5), A005917 (n=6), A348211 (condensed g.f.)

A:= func< n,k | (&+[(-1)^(j+1)*Binomial(n,j)*Binomial((n-2*j)*k+n-j-2,n-3)/2 : j in [0..Floor((n-1)/2)]]) >;
A348210:= func< n,k | A(n-k,k) >;
[A348210(n,k): k in [0..n-2], n in [2..13]]; // G. C. Greubel, Feb 28 2024

A348210 := proc(n,k)
    local a,j ;
    a := 0 ;
    for j from 0 to floor((n-1)/2) do
            a := a+ (-1)^j *binomial(n,j) *binomial( (n-2*j)*k+n-j-2,n-3) ;
    end do:
    -a/2 ;
end proc:
seq( seq( A348210(d-k,k),k=0..d-2),d=2..12) ;

A[n_, k_] := (-1/2)*Sum[(-1)^j*Binomial[n, j]*Binomial[(n - 2*j)*k + n - j - 2, n - 3], {j, 0, Floor[(n - 1)/2]}];
Table[A[n - k, k], {n, 2, 13}, {k, 0, n - 2}] // Flatten (* Jean-François Alcover, Mar 06 2023 *)

def A(n,k): return sum( (-1)^(j+1)*binomial(n,j)*binomial((n-2*j)*k+n-j-2,n-3) for j in range(1+(n-1)//2) )/2
def A348210(n,k): return A(n-k, k)
flatten([[A348210(n,k) for k in range(n-1)] for n in range(2,13)]) # G. C. Greubel, Feb 28 2024

A(n,k) = (-1/2)*Sum_{j=0..floor((n-1)/2)} (-1)^j *binomial(n,j) *binomial((n-2*j)*k+n-j-2,n-3).
A(7,k) = 1 + 7*k*(k+1)*(11*k^2+11*k+8)/12.
A(8,k) = (2*k+1)*(4*k^2+6*k+3)*(4*k^2+2*k+1)/3.
A(9,k) = 1 + k*(k+1)*(289*k^4+578*k^3+581*k^2+292*k+108)/16.

A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 40, 31, 9, 1, 1, 21, 85, 105, 51, 11, 1, 1, 28, 161, 295, 219, 76, 13, 1, 1, 36, 280, 721, 771, 396, 106, 15, 1, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1

Offset: 0

Peter Bala, Oct 31 2008

Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.

This seems to be a duplicate of A273350. - Alois P. Heinz, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - N. J. A. Sloane, Jul 02 2016

			The array begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1.....1.....1.....1.....1.....1
1..|..1.....3.....5.....7.....9....11
2..|..1.....6....16....31....51....76
3..|..1....10....40...105...219...396
4..|..1....15....85...295...771..1681
5..|..1....21...161...721..2331..6083
...
Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .
Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .

Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).

Cf. also A273350.

Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jan 15 2018 *)

taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);
T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* Vladimir Kruchinin, Jan 15 2018 */

Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.
Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).
The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).
Contribution from Paul Barry, Jan 06 2009: (Start)
The g.f. for the corresponding number triangle is:
1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)
This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - R. J. Mathar, Jun 16 2016
G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018
T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - Vladimir Kruchinin, Jan 15 2018

A133285 Indices of the centered 12-gonal numbers which are also 12-gonal number, or numbers X such that 120X^2-120X+36 is a square.

Original entry on oeis.org

1, 12, 253, 5544, 121705, 2671956, 58661317, 1287877008, 28274632849, 620754045660, 13628314371661, 299202162130872, 6568819252507513, 144214821393034404, 3166157251394249365, 69511244709280451616

Offset: 1

Richard Choulet, Oct 16 2007

Partial sums of A077422. - R. J. Mathar, Nov 27 2011

Indices of centered pentagonal numbers (A005891) which are also centered hexagonal numbers (A003215). - Colin Barker, Feb 07 2015

Colin Barker, Table of n, a(n) for n = 1..746
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Cf. A003215, A005891, A077422, A133141, A254782.

Table[Cos[n*ArcCos[11]] // Round, {n, 0, 15}] // Accumulate  (* Jean-François Alcover, Dec 19 2013, after R. J. Mathar *)
LinearRecurrence[{23,-23,1},{1,12,253},20] (* Harvey P. Dale, Jul 04 2018 *)

Vec(x*(11*x-1)/((x-1)*(x^2-22*x+1)) + O(x^100)) \\ Colin Barker, Feb 07 2015

a(n+2) = 22*a(n+1)-a(n)-10 ; a(n+1)=11*a(n)-5+(120*a(n)^2-120*a(n)+36)^0.5

G.f. x*(-1+11*x) / ( (x-1)*(x^2-22*x+1) ). - R. J. Mathar, Nov 27 2011

More terms from Paolo P. Lava, Aug 06 2008

A285810 Primes equal to a centered pentagonal number plus 1.

Original entry on oeis.org

2, 7, 17, 107, 227, 277, 457, 857, 1627, 3517, 4517, 5407, 9767, 11057, 13877, 15017, 16607, 20477, 23767, 26267, 27827, 35107, 37517, 41927, 42577, 50767, 53657, 58907, 62017, 68477, 79657, 83267, 86027, 93607, 98507, 110777, 113957, 128257, 137477, 145807

Offset: 1

Colin Barker, Apr 27 2017

nonn

Colin Barker, Table of n, a(n) for n = 1..1000

Cf. A000040, A005891, A285809, A285811, A285812.

cpg(m, n) = m*n*(n-1)/2+1 \\ n-th centered m-gonal number
maxk=600; L=List(); for(k=1, maxk, if(isprime(p=cpg(5, k) + 1), listput(L, p))); Vec(L)

A133272 Indices of centered heptagonal numbers (A069099) which are also heptagonal numbers (A000566).

Original entry on oeis.org

1, 7, 78, 924, 11005, 131131, 1562562, 18619608, 221872729, 2643853135, 31504364886, 375408525492, 4473397941013, 53305366766659, 635191003258890, 7568986672340016, 90192649064821297, 1074742802105515543

Offset: 1

Richard Choulet, Oct 16 2007

Numbers X such that 140*X^2-140*X+49 is a square.
Also positive integers x in the solutions to 5*x^2 - 7*y^2 - 5*x + 7*y = 0, the corresponding values of y being A253621. - Colin Barker, Jan 06 2015
Also indices of centered pentagonal numbers (A005891) which are also centered heptagonal numbers (A069099). - Colin Barker, Jan 06 2015

Colin Barker, Table of n, a(n) for n = 1..930
Index entries for linear recurrences with constant coefficients, signature (13,-13,1).

Cf. A000566, A069099, A128919, A253621, A253622.

LinearRecurrence[{13,-13,1},{1,7,78},25] (* Paolo Xausa, Jan 07 2024 *)

Vec(x*(6*x-1)/((x-1)*(x^2-12*x+1)) + O(x^100)) \\ Colin Barker, Jan 06 2015

a(n+2) = 12*a(n+1) - a(n) - 5.
a(n+1) = 6*a(n) - 5/2 + (1/2)*sqrt(140*a(n)^2 - 140*a(n) + 49).
G.f.: x*(-1+6*x)/((-1+x)*(1-12*x+x^2)). - R. J. Mathar, Nov 14 2007
a(n) = 13*a(n-1) - 13*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015

More terms from Paolo P. Lava, Jul 14 2008

A133273 Indices of centered decagonal numbers which are also decagonal numbers.

Original entry on oeis.org

1, 10, 171, 3060, 54901, 985150, 17677791, 317215080, 5692193641, 102142270450, 1832868674451, 32889493869660, 590178020979421, 10590314883759910, 190035489886698951, 3410048503076821200, 61190837565496082641, 1098025027675852666330, 19703259660599851911291

Offset: 1

Richard Choulet, Oct 16 2007

Numbers k such that 80*k^2 - 80*k + 25 is a square.

Also the indices of centered square numbers which are also centered pentagonal numbers. - Colin Barker, Jan 01 2015

Colin Barker, Table of n, a(n) for n = 1..798
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A001107, A001844, A005891, A010532, A062786, A128922.

LinearRecurrence[{19,-19,1},{1,10,171},20] (* Harvey P. Dale, Oct 09 2020 *)

Vec(x*(-1+9*x)/((-1+x)*(1-18*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n+2) = 18*a(n+1) - a(n) - 8.
a(n+1) = 9*a(n) - 4 + sqrt(80*a(n)^2 - 80*a(n) + 25).
G.f.: x*(-1+9*x)/(-1+x)/(1 - 18*x + x^2). - R. J. Mathar, Nov 14 2007
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3). - Colin Barker, Jan 01 2015
Product_{n>=2} (1 - 1/a(n)) = 2/sqrt(5) (= A010532 / 10). - Amiram Eldar, Dec 02 2024

More terms from Paolo P. Lava, Nov 25 2008

cp's OEIS Frontend

A280952 Expansion of Product_{k>=0} 1/(1 - x^(5*k*(k+1)/2+1)).

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A069128 Centered 15-gonal numbers: a(n) = (15*n^2 - 15*n + 2)/2.

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A212656 a(n) = 5*n^2 + 1.

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A281083 Expansion of Product_{k>=0} (1 + x^(5*k*(k+1)/2+1)).

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A348210 Varma's Kosta numbers of semi-standard tableaux: array A(n>=2, k>=0) read by rising antidiagonals.

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A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

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A133285 Indices of the centered 12-gonal numbers which are also 12-gonal number, or numbers X such that 120*X^2-120*X+36 is a square.

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A280952 Expansion of Product_{k>=0} 1/(1 - x^(5k(k+1)/2+1)).

A069128 Centered 15-gonal numbers: a(n) = (15n^2 - 15n + 2)/2.

A281083 Expansion of Product_{k>=0} (1 + x^(5k(k+1)/2+1)).

A133285 Indices of the centered 12-gonal numbers which are also 12-gonal number, or numbers X such that 120X^2-120X+36 is a square.