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 1-3 of 3 results.

A060047 Triangle of generalized sum of divisors function, read by rows.

Original entry on oeis.org

1, 2, 4, 1, 4, 2, 6, 4, 8, 8, 8, 14, 8, 1, 18, 13, 2, 28, 12, 4, 40, 12, 8, 52, 16, 14, 70, 14, 24, 88, 16, 40, 104, 24, 1, 56, 140, 16, 2, 84, 168, 18, 4, 122, 196, 26, 8, 168, 240, 20, 14, 232, 278, 24, 24, 312, 320, 32, 40, 408, 380, 24, 64, 528, 440, 24, 100, 672, 504
Offset: 1

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Author

N. J. A. Sloane, Mar 19 2001

Keywords

Comments

Lengths of rows are 1 1 1 2 2 2 2 2 3 3 3 3 3 3 3 ... (A000196).

Examples

			Triangle turned on its side begins:
  1  2  4  4  6  8  8  8 13 12 12 ...
           1  2  4  8 14 18 28 40 ...
                          1  2  4 ...
For example, T(6,1) = 8, T(6,2) = 4.
		

Crossrefs

Formula

T(n, k) = sum of s_1*s_2*...*s_k where s_1, s_2, ..., s_k are such that s_1*(2*m_1-1) + s_2*(2*m_2-1) + ... + s_k*(2*m_k-1) = n and the sum is over all such k-partitions of n.
G.f. for k-th diagonal (the k-th row of the sideways triangle shown in the example): Sum_{ m_1 < m_2 < ... < m_k} q^(2*m_1+2*m_2+...+2*m_k-k)/((1-q^{2*m_1-1})*(1-q^{2*m_2-1})*...*(1-q^{2*m_k-1}))^2 = Sum_n T(n, k)*q^n.
G.f. for k-th diagonal: (-1)^k * (1/k) * ( Sum_{j>=k} (-1)^j * j * binomial(j+k-1,2*k-1) * q^(j^2) ) / ( 1 + 2 * Sum_{j>=1} (-q)^(j^2) ). - Seiichi Manyama, Sep 15 2023

Extensions

More terms from Naohiro Nomoto, Jan 24 2002

A365665 Expansion of Sum_{0

Original entry on oeis.org

1, 3, 9, 22, 51, 108, 208, 390, 693, 1193, 1977, 3195, 4995, 7722, 11583, 17164, 24882, 35685, 50205, 70083, 96300, 131101, 176358, 235377, 310651, 407352, 529074, 682750, 874038, 1112085, 1405521, 1766259, 2206413, 2741431, 3389052, 4168089, 5103450, 6218469
Offset: 15

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Author

Seiichi Manyama, Sep 15 2023

Keywords

Comments

Number of partitions of n with five designated summands (when part i has multiplicity j > 0 exactly one part i is "designated"). For example: a(16) = 3 because there are three partitions of 16 with five designated summands: [6'+ 4'+ 3'+ 2'+ 1'], [5'+ 4'+ 3'+ 2'+ 1'+ 1], [5'+ 4'+ 3'+ 2'+ 1 + 1']. - Omar E. Pol, Jul 29 2025

Crossrefs

A diagonal of A060043.
Column k=5 of A385001.
Cf. A384926.

Programs

  • Mathematica
    nmax = 60; Drop[CoefficientList[Series[-1/11 * Sum[(-1)^k*(2*k + 1)*Binomial[k + 5, 10]*x^(k*(k + 1)/2), {k, 5, nmax}]/Sum[(-1)^k*(2*k + 1)*x^(k*(k + 1)/2), {k, 0, nmax}], {x, 0, nmax}], x], 15] (* Vaclav Kotesovec, Jul 29 2025 *)
    (* or *)
    Table[(10679/17203200 - 1571*n/774144 + 133*n^2/92160 - n^3/3072 + n^4/46080) * DivisorSigma[1, n] + (1571/1548288 - 133*n/122880 + 3*n^2/10240 - n^3/46080) * DivisorSigma[3, n] + (133/1228800 - n/20480 + n^2/215040) * DivisorSigma[5, n] + (1/516096 - n/3096576) * DivisorSigma[7, n] + DivisorSigma[9, n]/154828800, {n, 15, 60}] (* Vaclav Kotesovec, Jul 29 2025 *)

Formula

G.f.: -(1/11) * ( Sum_{k>=5} (-1)^k * (2*k+1) * binomial(k+5,10) * q^(k*(k+1)/2) ) / ( Sum_{k>=0} (-1)^k * (2*k+1) * q^(k*(k+1)/2) ).
From Vaclav Kotesovec, Jul 29 2025: (Start)
a(n) = (10679/17203200 - 1571*n/774144 + 133*n^2/92160 - n^3/3072 + n^4/46080)*sigma(n) + (1571/1548288 - 133*n/122880 + 3*n^2/10240 - n^3/46080)*sigma_3(n) + (133/1228800 - n/20480 + n^2/215040)*sigma_5(n) + (1/516096 - n/3096576)*sigma_7(n) + sigma_9(n)/154828800.
Sum_{k=1..n} a(k) ~ Pi^10 * n^10 / 144850083840000.
(End)

A365666 Expansion of Sum_{0

Original entry on oeis.org

1, 2, 4, 8, 14, 24, 40, 64, 100, 144, 212, 304, 424, 588, 800, 1072, 1422, 1864, 2408, 3080, 3950, 4972, 6224, 7760, 9564, 11742, 14344, 17384, 20968, 25204, 30112, 35840, 42548, 50078, 58888, 69048, 80474, 93628, 108608, 125408, 144536, 166224, 190348
Offset: 16

Views

Author

Seiichi Manyama, Sep 15 2023

Keywords

Crossrefs

A diagonal of A060047.
Cf. A015128.

Programs

  • Mathematica
    nmax = 60; Drop[CoefficientList[Series[1/4 * Sum[(-1)^k*k*Binomial[k + 3, 7]*x^(k^2), {k, 4, nmax}]/(1 + 2*Sum[(-x)^(k^2), {k, 1, nmax}]), {x, 0, nmax}], x], 16] (* Vaclav Kotesovec, Jul 30 2025 *)

Formula

G.f.: (1/4) * ( Sum_{k>=4} (-1)^k * k * binomial(k+3,7) * q^(k^2) ) / ( 1 + 2 * Sum_{k>=1} (-q)^(k^2) ).
Showing 1-3 of 3 results.