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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A351008 Alternately strict partitions. Number of even-length integer partitions y of n such that y_i > y_{i+1} for all odd i.

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 19, 23, 28, 34, 41, 50, 60, 71, 85, 102, 120, 142, 168, 197, 231, 271, 316, 369, 429, 497, 577, 668, 770, 888, 1023, 1175, 1348, 1545, 1767, 2020, 2306, 2626, 2990, 3401, 3860, 4379, 4963, 5616, 6350, 7173, 8093
Offset: 0

Views

Author

Gus Wiseman, Jan 31 2022

Keywords

Comments

Write the series in the g.f. given below as Sum_{k >= 0} q^(1 + 3 + 5 + ... + 2*k-1 + 2*k)/Product_{i = 1..2*k} 1 - q^i. Since 1/Product_{i = 1..2*k} 1 - q^i is the g.f. for partitions with parts <= 2*k, we see that the k-th summand of the series is the g.f. for partitions with largest part 2*k in which every odd number less than 2*k appears at least once as a part. The partitions of this type are conjugate to (and hence equinumerous with) the partitions (y_1, y_2, ..., y_{2*k}) of even length 2*k having strict decrease y_i > y_(i+1) for all odd i < 2*k. - Peter Bala, Jan 02 2024

Examples

			The a(3) = 1 through a(13) = 12 partitions (A..C = 10..12):
  21   31   32   42   43   53     54     64     65     75     76
            41   51   52   62     63     73     74     84     85
                      61   71     72     82     83     93     94
                           3221   81     91     92     A2     A3
                                  4221   4321   A1     B1     B2
                                         5221   4331   4332   C1
                                                5321   5331   5332
                                                6221   5421   5431
                                                       6321   6331
                                                       7221   6421
                                                              7321
                                                              8221
a(10) = 6: the six partitions 64, 73, 82, 91, 4321 and 5221 listed above have conjugate partitions 222211, 2221111, 22111111, 211111111, 4321 and 43111, These are the partitions of 10 with largest part L even and such that every odd number less than L appears at least once as a part. - _Peter Bala_, Jan 02 2024
		

Crossrefs

The version for equal instead of unequal is A035363.
The alternately equal and unequal version is A035457, any length A351005.
This is the even-length case of A122129, opposite A122135.
The odd-length version appears to be A122130.
The alternately unequal and equal version is A351007, any length A351006.

Programs

  • Maple
    series(add(q^(n*(n+2))/mul(1 - q^k, k = 1..2*n), n = 0..10), q, 141):
    seq(coeftayl(%, q = 0, n), n = 0..140); # Peter Bala, Jan 03 2025
  • Mathematica
    Table[Length[Select[IntegerPartitions[n],EvenQ[Length[#]]&&And@@Table[#[[i]]!=#[[i+1]],{i,1,Length[#]-1,2}]&]],{n,0,30}]

Formula

Conjecture: a(n+1) = A122129(n+1) - A122130(n). - Gus Wiseman, Feb 21 2022
G.f.: Sum_{n >= 0} q^(n*(n+2))/Product_{k = 1..2*n} 1 - q^k = 1 + q^3 + q^4 + 2*q^5 + 2*q^6 + 3*q^7 + 4*q^8 + 5*q^9 + 6*q^10 + .... - Peter Bala, Jan 02 2024

A356603 Position in A356226 of first appearance of the n-th composition in standard order (row n of A066099).

Original entry on oeis.org

1, 2, 4, 10, 8, 20, 50, 110, 16, 40, 100, 220, 250, 550, 1210, 1870, 32, 80, 200, 440, 500, 1100, 2420, 3740, 1250, 2750, 6050, 9350, 13310, 20570, 31790, 43010, 64, 160, 400, 880, 1000, 2200, 4840, 7480, 2500, 5500, 12100, 18700, 26620, 41140, 63580, 86020
Offset: 0

Views

Author

Gus Wiseman, Aug 30 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
The image consists of all numbers whose prime indices are odd and cover an initial interval of odd positive integers.

Examples

			The terms together with their prime indices begin:
      1: {}
      2: {1}
      4: {1,1}
     10: {1,3}
      8: {1,1,1}
     20: {1,1,3}
     50: {1,3,3}
    110: {1,3,5}
     16: {1,1,1,1}
     40: {1,1,1,3}
    100: {1,1,3,3}
    220: {1,1,3,5}
    250: {1,3,3,3}
    550: {1,3,3,5}
   1210: {1,3,5,5}
   1870: {1,3,5,7}
		

Crossrefs

See link for sequences related to standard compositions.
The partitions with these Heinz numbers are counted by A053251.
A subset of A066208 (numbers with all odd prime indices).
Up to permutation, these are the positions of first appearances of rows in A356226. Other statistics are:
- length: A287170, firsts A066205
- minimum: A356227
- maximum: A356228
- bisected length: A356229
- standard composition: A356230
- Heinz number: A356231
The sorted version is A356232.
An ordered version is counted by A356604.
A001221 counts distinct prime factors, sum A001414.
A073491 lists numbers with gapless prime indices, complement A073492.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    stcinv[q_]:=1/2 Total[2^Accumulate[Reverse[q]]];
    mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
    sq=stcinv/@Table[Length/@Split[primeMS[n],#1>=#2-1&],{n,1000}];
    Table[Position[sq,k][[1,1]],{k,0,mnrm[Rest[sq]]}]

A357624 Skew-alternating sum of the reversed n-th composition in standard order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The 357-th composition is (2,1,3,2,1) so a(357) = 1 - 2 - 3 + 2 + 1 = -1.
The 358-th composition is (2,1,3,1,2) so a(358) = 2 - 1 - 3 + 1 + 2 = 1.
		

Crossrefs

See link for sequences related to standard compositions.
The half-alternating form is A357622, non-reverse A357621.
The reverse version is A357623.
Positions of zeros are A357628, non-reverse A357627.
The version for prime indices is A357630.
The version for Heinz numbers of partitions is A357634.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[skats[Reverse[stc[n]]],{n,0,100}]

A356841 Numbers k such that the k-th composition in standard order covers an interval of positive integers (gapless).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 18, 20, 21, 22, 23, 26, 27, 29, 30, 31, 32, 36, 37, 38, 41, 42, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 64, 68, 72, 74, 75, 77, 78, 82, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101
Offset: 1

Views

Author

Gus Wiseman, Aug 31 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms and their corresponding standard compositions begin:
   0: ()
   1: (1)
   2: (2)
   3: (1,1)
   4: (3)
   5: (2,1)
   6: (1,2)
   7: (1,1,1)
   8: (4)
  10: (2,2)
  11: (2,1,1)
  13: (1,2,1)
  14: (1,1,2)
  15: (1,1,1,1)
  16: (5)
  18: (3,2)
  20: (2,3)
  21: (2,2,1)
		

Crossrefs

See link for sequences related to standard compositions.
An unordered version is A073491, complement A073492.
These compositions are counted by A107428.
The complement is A356842.
The non-initial case is A356843, unordered A356845.
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.
A356844 ranks compositions with at least one 1.

Programs

  • Mathematica
    nogapQ[m_]:=m=={}||Union[m]==Range[Min[m],Max[m]];
    stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],nogapQ[stc[#]]&]

A357635 Numbers k such that the half-alternating sum of the partition having Heinz number k is 1.

Original entry on oeis.org

2, 8, 24, 32, 54, 128, 135, 162, 375, 384, 512, 648, 864, 875, 1250, 1715, 1944, 2048, 2160, 2592, 3773, 4374, 4802, 5000, 6000, 6144, 8192, 9317, 10368, 10935, 13122, 13824, 14000, 15000, 17303, 19208, 20000, 24167, 27440, 29282, 30375, 31104, 32768, 33750
Offset: 1

Views

Author

Gus Wiseman, Oct 28 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

Examples

			The terms together with their prime indices begin:
    2: {1}
    8: {1,1,1}
   24: {1,1,1,2}
   32: {1,1,1,1,1}
   54: {1,2,2,2}
  128: {1,1,1,1,1,1,1}
  135: {2,2,2,3}
  162: {1,2,2,2,2}
  375: {2,3,3,3}
  384: {1,1,1,1,1,1,1,2}
  512: {1,1,1,1,1,1,1,1,1}
  648: {1,1,1,2,2,2,2}
  864: {1,1,1,1,1,2,2,2}
  875: {3,3,3,4}
		

Crossrefs

The version for k = 0 is A000583, standard compositions A357625-A357626.
The version for original alternating sum is A345958.
Positions of ones in A357633, non-reverse A357629.
The skew version for k = 0 is A357636, non-reverse A357632.
These partitions are counted by A035444, skew A035544.
The non-reverse version is A357851, k = 0 version A357631.
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, even-length A357642.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Select[Range[1000],halfats[Reverse[primeMS[#]]]==1&]

A053252 Coefficients of the '3rd-order' mock theta function chi(q).

Original entry on oeis.org

1, 1, 1, 0, 0, 0, 1, 1, 0, 0, -1, 0, 1, 1, 1, -1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 1, 1, 0, -1, -1, 1, 1, 0, -1, -1, 0, 1, 2, 1, -1, -1, 0, 1, 1, 0, -1, -2, 0, 1, 2, 1, -1, -1, -1, 1, 2, 1, -1, -2, -1, 2, 2, 1, -1, -2, -1, 1, 2, 0, -1, -3, 0, 2, 3, 2, -2, -2, -1, 2, 3, 0, -2, -3, -1, 2, 3, 2, -3, -3, -1, 2, 4, 1, -2, -4, -1, 3, 4, 2, -2, -4
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

References

  • N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 55, Eq. (26.14).
  • Srinivasa Ramanujan, Collected Papers, Chelsea, New York, 1962, pp. 354-355.
  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 17.

Crossrefs

Other '3rd-order' mock theta functions are at A000025, A053250, A053251, A053253, A053254, A053255, A261401.

Programs

  • Mathematica
    Series[Sum[q^n^2/Product[1-q^k+q^(2k), {k, 1, n}], {n, 0, 10}], {q, 0, 100}]

Formula

G.f.: chi(q) = Sum_{n >= 0} q^n^2/((1-q+q^2)*(1-q^2+q^4)*...*(1-q^n+q^(2n))).
G.f.: G(0), where G(k) = 1 + q^(k+1) / (1 - q^(k+1)) / G(k+1). - Joerg Arndt, Jun 29 2013

A053255 Coefficients of the '3rd-order' mock theta function rho(q).

Original entry on oeis.org

1, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, 0, 2, -1, -1, 1, -1, -1, 2, -1, 0, 2, -1, -1, 2, -2, -1, 3, -2, -1, 3, -2, -1, 3, -2, -1, 4, -3, -1, 4, -2, -2, 4, -3, -2, 5, -4, -2, 6, -3, -2, 6, -4, -2, 7, -5, -2, 7, -5, -3, 8, -6, -3, 9, -6, -3, 10, -6, -4, 10, -7, -4, 12, -8, -4, 13, -8, -5, 13, -9, -5, 15, -10, -5, 16, -11, -6, 17, -12, -7, 19, -13, -6, 21, -13
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

References

  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 15.

Crossrefs

Other '3rd-order' mock theta functions are at A000025, A053250, A053251, A053252, A053253, A053254.

Programs

  • Mathematica
    Series[Sum[q^(2n(n+1))/Product[1+q^(2k+1)+q^(4k+2), {k, 0, n}], {n, 0, 6}], {q, 0, 100}]

Formula

G.f.: rho(q) = Sum_{n >= 0} q^(2*n*(n+1))/((1+q+q^2)*(1+q^3+q^6)*...*(1+q^(2*n+1)+q^(4*n+2))).

A122130 Expansion of f(-x^4, -x^16) / psi(-x) in powers of x where psi() is a Ramanujan theta function and f(, ) is Ramanujan's general theta function.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 11, 14, 18, 22, 27, 34, 41, 50, 61, 73, 88, 106, 126, 150, 179, 211, 249, 294, 345, 404, 473, 551, 642, 747, 865, 1002, 1159, 1336, 1539, 1771, 2033, 2331, 2670, 3052, 3485, 3976, 4527, 5150, 5854, 6642, 7530, 8529, 9647, 10902
Offset: 0

Views

Author

Michael Somos, Aug 21 2006, corrected Aug 21 2006

Keywords

Comments

Generating function arises naturally in Rodney Baxter's solution of the Hard Hexagon Model according to George Andrews.
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
From Gus Wiseman, Feb 19 2022: (Start)
This appears to be the number of odd-length alternately strict integer partitions of n + 1, i.e., partitions y such that y_i != y_{i+1} for all odd i. For example, the a(1) = 1 through a(9) = 7 partitions are:
(1) (2) (3) (4) (5) (6) (7) (8) (9)
(211) (311) (321) (322) (422) (432)
(411) (421) (431) (522)
(511) (521) (531)
(611) (621)
(711)
(32211)
The even-length version is A351008. Including even-length partitions appears to give A122129. Swapping strictly and weakly decreasing relations gives A351595. The constant instead of strict version is A351594. (End)
Wiseman's first conjecture above was proved by Connor, Proposition 2. - Peter Bala, Dec 22 2024

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 5*x^7 + 7*x^8 + 9*x^9 + ...
G.f. = q^31 + q^71 + q^111 + 2*q^151 + 2*q^191 + 3*q^231 + 4*q^271 + 5*q^311 + ...
		

References

  • G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1999; Exercise 6(b), p. 591.
  • G. E. Andrews, q-series, CBMS Regional Conference Series in Mathematics, 66, Amer. Math. Soc. 1986, see p. 8, Eq. (1.8). MR0858826 (88b:11063)

Crossrefs

Programs

  • Mathematica
    nmax = 100; CoefficientList[Series[Product[1/((1-x^(2*k-1))*(1-x^(20*k-8))*(1-x^(20*k-12))), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 30 2015 *)
    a[ n_] := SeriesCoefficient[ 1 / (QPochhammer[x, x^2] QPochhammer[x^8, x^20] QPochhammer[x^12, x^20]), {x, 0, n}]; (* Michael Somos, Nov 12 2016 *)
    a[ n_] := SeriesCoefficient[ Sqrt[2] x^(1/8) QPochhammer[ x^4, x^20] QPochhammer[ x^16, x^20] QPochhammer[x^20] / EllipticTheta[ 2, Pi/4, x^(1/2)], {x, 0, n}] // Simplify; (* Michael Somos, Nov 12 2016 *)
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( sum(k=1, sqrtint(n+1), x^(k^2-1) / prod(i=1, 2*k-1, 1 - x^i, 1 + x * O(x^(n-k^2+1)))), n))};

Formula

Expansion of f(x, x^9) / f(-x^2, -x^3) in powers of x where f(, ) is Ramanujan's general theta function. - Michael Somos, Nov 12 2016
Expansion of f(-x^2) * f(-x^20) / (f(-x) * f(-x^8, -x^12)) in powers of x where f(-x) : = f(-x, -x^2) and f(, ) is Ramanujan's general theta function.
Euler transform of period 20 sequence [ 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, ...].
G.f.: Sum_{k>0} x^(k^2 - 1) / ((1 - x) * (1 - x^2) * ... * (1 - x^(2k-1))).
G.f.: 1/(Product_{k>0} (1-x^(2k-1))(1-x^(20k-8))(1-x^(20k-12))).
a(n) ~ (3-sqrt(5))^(1/4) * exp(Pi*sqrt(2*n/5)) / (4*sqrt(5)*n^(3/4)). - Vaclav Kotesovec, Aug 30 2015

A357704 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 2, 0, 3, 0, 0, 2, 2, 0, 3, 0, 0, 3, 1, 3, 0, 4, 0, 0, 3, 2, 4, 2, 0, 4, 0, 0, 4, 2, 6, 2, 3, 0, 5, 0, 0, 4, 3, 5, 7, 3, 3, 0, 5, 0, 0, 5, 3, 8, 4, 10, 2, 4, 0, 6, 0, 0, 5, 4, 8, 6, 11, 9, 3, 4, 0, 6, 0, 0, 6, 4, 11, 5, 15, 8, 13, 3, 5, 0, 7
Offset: 0

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

Examples

			Triangle begins:
  1
  0  1
  0  0  2
  0  0  1  2
  0  0  2  0  3
  0  0  2  2  0  3
  0  0  3  1  3  0  4
  0  0  3  2  4  2  0  4
  0  0  4  2  6  2  3  0  5
  0  0  4  3  5  7  3  3  0  5
  0  0  5  3  8  4 10  2  4  0  6
  0  0  5  4  8  6 11  9  3  4  0  6
  0  0  6  4 11  5 15  8 13  3  5  0  7
  0  0  6  5 11  8 13 19 10 13  4  5  0  7
  0  0  7  5 14  8 19 13 25  9 17  4  6  0  8
  0  0  7  6 14 11 19 17 29 23 13 18  5  6  0  8
Row n = 7 counts the following reversed partitions:
  .  .  (115)   (124)   (133)      (11113)   .  (7)
        (1114)  (1222)  (223)      (111112)     (16)
        (1123)          (11122)                 (25)
                        (1111111)               (34)
		

Crossrefs

Row sums are A000041.
Last entry of row n is A008619(n).
The central column in the non-reverse case is A035363, skew A035544.
For original reverse-alternating sum we have A344612.
For original alternating sum we have A344651, ordered A097805.
The non-reverse version is A357637, skew A357638.
The central column is A357639, skew A357640.
The non-reverse ordered version (compositions) is A357645, skew A357646.
The skew-alternating version is A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[Length[Select[Reverse/@IntegerPartitions[n],halfats[#]==k&]],{n,0,15},{k,-n,n,2}]

A003475 Expansion of Sum_{k>0} (-1)^(k+1) q^(k^2) / ((1-q)(1-q^3)(1-q^5)...(1-q^(2k-1))).

Original entry on oeis.org

1, 1, 1, 0, 0, 0, -1, -1, 0, -1, -1, 0, -1, 0, 1, -1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, -1, 0, 0, 0, 1, -1, -1, -1, 0, 0, -1, 0, -1, 0, 0, -1, -1, -1, 1, 0, -1, 0, 0, 0, 1, 0, -1, 0, 0, 1, 0, 1, 1, 0, 1, -1, 1, 0, 0, 2, 0, 0, 0, 1, 0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 1, -1, -1, 1, 0, 0, 0, -1, -2, 0, 0, -1, 0, 1, 0, -1, -1, -1, 0, 0, 0
Offset: 1

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Comments

|a(n)|<3 if n<1036, a(1036)=3. - Michael Somos, Sep 16 2006

Examples

			G.f. = x + x^2 + x^3 - x^7 - x^8 - x^10 - x^11 - x^13 + x^15 - x^16 + ...
		

References

  • F. J. Dyson, A walk through Ramanujan's garden, pp. 7-28 of G. E. Andrews et al., editors, Ramanujan Revisited. Academic Press, NY, 1988.
  • F. J. Dyson, Selected Papers, Am. Math. Soc., 1996, p. 204.

Crossrefs

Programs

  • Maple
    P:=n->mul((1-q^(2*i+1)),i=0..n-1):
    t5:=add((-1)^(n+1)*q^(n^2)/P(n),n=1..40):
    t6:=series(t5,q,40); # Based on Patkowski, 2010, Eq. 3.1. - N. J. A. Sloane, Jun 29 2011
  • Mathematica
    a[ n_] := SeriesCoefficient[ 1 - QHypergeometricPFQ[ {x^2}, {x}, x^2, x], {x, 0, n}]; (* Michael Somos, Feb 02 2015 *)
  • PARI
    {a(n) = local(A, p, e, x, y); if( n<0, 0, n = 24*n-1; A = factor(n); prod(k=1, matsize(A)[1], if( p=A[k, 1], e=A[k, 2]; if( p<5, 0, if( p%24>1 && p%24<23, if( e%2, 0, if( p%24==7 || p%24==17, (-1)^(e/2), 1)), x=y=0; if( p%24==1, forstep(i=1, sqrtint(p), 2, if( issquare( (i^2 + p) / 2, &y), x=i; break)), for(i=1, sqrtint(p\2), if( issquare( 2*i^2 + p,&x), y=i; break))); (e+1) * (-1)^( (x + if((x-y)%6, y, -y)) / 6*e))))) / -2)}; /* Michael Somos, Aug 17 2006 */
    
  • PARI
    {a(n) = local(A); if( n<1, 0, A = -1 + x * O(x^n); polcoeff( sum(k=1, sqrtint(n), A *= 1 / (1 - x^(1 - 2*k)) * (1 + x * O(x^(n - k^2)))), n))}; /* Michael Somos, Sep 16 2006 */

Formula

Define c(24*k + 1) = A003406(k), c(24*k - 1) = -2*A003475(k), c(n) = 0 otherwise. Then c(n) is multiplicative with c(2^e) = c(3^e) = 0^e, c(p^e) = (-1)^(e/2) * (1 + (-1)^e) / 2 if p == 7, 17 (mod 24), c(p^e) = (1 + (-1)^e) / 2 if p == 5, 11, 13, 19 (mod 24), c(p^e) = (e+1)*(-1)^(y*e) where p == 1, 23 (mod 24) and p = x^2 - 72*y^2 . - Michael Somos, Aug 17 2006
G.f.: x + x^2 * (1 - x^2) + x^3 * (1 - x^2) * (1 - x^4) + x^4 * (1 - x^2) * (1 - x^4) * (1 - x^6) + ... . - Michael Somos, Aug 18 2006
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