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 A026810 Number of partitions of n in which the greatest part is 4. 33
 0, 0, 0, 0, 1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169, 185, 206, 225, 249, 270, 297, 321, 351, 378, 411, 441, 478, 511, 551, 588, 632, 672, 720, 764, 816, 864, 920, 972, 1033, 1089, 1154, 1215, 1285, 1350 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,7 COMMENTS Also number of partitions of n into exactly 4 parts. Also the number of weighted cubic graphs on 4 nodes (=the tetrahedron) with weight n. - R. J. Mathar, Nov 03 2018 REFERENCES G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275. D. E. Knuth, The Art of Computer Programming, vol. 4,fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31. LINKS Washington Bomfim, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1). FORMULA G.f.: x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) = x^4/((1-x)^4*(1+x)^2*(1+x+x^2)*(1+x^2)). a(n+4) = A001400(n). - Michael Somos, Apr 07 2012 a(n) = round(n^3 +3*n^2 -9*n[n odd])/144, where [...] denotes the Iverson bracket [true] = 1 and [false] = 0. - Washington Bomfim, Jul 03 2012 a(n) = (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 -A049347(n)/9 +A056594(n)/8. - R. J. Mathar, Jul 03 2012 From Gregory L. Simay, Oct 13 2015: (Start) a(n) = (n^3 + 3*n^2 - 9*n)/144 + a(m) - (m^3 + 3*m^2 - 9*m)/144 if n = 12k + m and m is odd. For example, a(23) = a(12*1 + 11) = (23^3 + 3*23^2 - 9*23)/144 + a(11) - (11^3 + 3*11^2 - 9*11)/144 = 94. a(n) = (n^3 + 3*n^2)/144 + a(m) - (m^3 + 3*m^2)/144 if n = 12k + m and m is even. For example, a(22) = a(12*1 + 10) = (22^3 + 3*22^2)/144 + a(10) - (10^3 + 3*10^2)/144 = 84. (End) a(n) = A008284(n,4). - Robert A. Russell, May 13 2018 a(2n+1) = a(2n) + a(n+1) - a(n-3) and a(2n) = a(2n-1) + a(n+2) - a(n-2). - Gregory L. Simay, Jul 28 2019 MAPLE A049347 := proc(n)         op(1+(n mod 3), [1, -1, 0]) ; end proc: A056594 := proc(n)         op(1+(n mod 4), [1, 0, -1, 0]) ; end proc: A026810 := proc(n)         1/288*(n+1)*(2*n^2+4*n-13+9*(-1)^n) ;         %-A049347(n)/9 ;         %+A056594(n)/8 ; end proc: # R. J. Mathar, Jul 03 2012 MATHEMATICA Table[Count[IntegerPartitions[n], {4, ___}], {n, 0, 60}] LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {0, 0, 0, 0, 1, 1, 2, 3, 5, 6}, 60] (* Vincenzo Librandi, Oct 14 2015 *) Table[Length[IntegerPartitions[n, {4}]], {n, 0, 60}] (* Eric Rowland, Mar 02 2017 *) CoefficientList[Series[x^4/Product[1 - x^k, {k, 1, 4}], {x, 0, 60}], x] (* Robert A. Russell, May 13 2018 *) PROG (PARI) for(n=0, 60, print(n, " ", round((n^3 + 3*n^2 -9*n*(n % 2))/144))); \\ Washington Bomfim, Jul 03 2012 (PARI) x='x+O('x^60); concat([0, 0, 0, 0], Vec(x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)))) \\ Altug Alkan, Oct 14 2015 (PARI) vector(60, n, n--; (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 + real(I^n)/8 - ((n+2)%3-1)/9) \\ Altug Alkan, Oct 26 2015 (PARI) print1(0, ", "); for(n=1, 60, j=0; forpart(v=n, j++, , [4, 4]); print1(j, ", ")) \\ Hugo Pfoertner, Oct 01 2018 (MAGMA) [Round((n^3+3*n^2-9*n*(n mod 2))/144): n in [0..60]]; // Vincenzo Librandi, Oct 14 2015 CROSSREFS Cf. A001400, A026811, A026812, A026813, A026814, A026815, A026816, A069905 (3 positive parts), A002621 (partial sums), A005044 (first differences). Sequence in context: A325863 A028309 A242717 * A001400 A008773 A008772 Adjacent sequences:  A026807 A026808 A026809 * A026811 A026812 A026813 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified October 14 02:29 EDT 2019. Contains 327995 sequences. (Running on oeis4.)