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 A078703 Number of ways of subtracting twice a triangular number from a perfect square to obtain the integer n. 16
 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 3, 1, 1, 2, 2, 2, 3, 1, 1, 2, 2, 2, 2, 1, 1, 4, 1, 2, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 1, 4, 1, 2, 3, 1, 2, 2, 1, 1, 4, 2, 1, 3, 2, 1, 4, 2, 1, 2, 1, 3, 3, 1, 2, 2, 2, 2, 2, 1, 1, 6, 2, 2, 2, 1, 2, 2, 2, 1, 4, 2, 1, 3, 1, 2, 4, 1, 1, 3, 2, 2, 4, 2, 2, 2, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Also number of symmetric unimodal consecutive integer sequences that sum to the integer n (e.g., 4+5+6+5+4 = 24 = n). Also number of double trapezoidal arrangements of n objects, denoted SDT(n); i.e., the number of ways to arrange n objects into symmetrically-placed, congruent isosceles trapezoids adjoined at overlapping largest bases. Also number of divisors of 4*n-1 of form 4*k+1 (or 4*k+3). - Vladeta Jovovic, Jan 05 2004. Therefore a(n) is one half of the number of divisors of A004767(n-1) (numbers 3 (mod 4)). - Wolfdieter Lang, Jul 29 2016 LINKS Seiichi Manyama, Table of n, a(n) for n = 1..10000 T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2 (1999), Article 99.1.6. FORMULA a(n) = ((r1 + 1)*(r2 + 1)*...*(rk + 1))/2, where ((p1^r1)*(p2^r2)*...*(pk^rk)) is the factorization of 4*n - 1 into (odd) primes. G.f.: Sum_{n>0} x^n/(1-x^(4*n-1)). - Vladeta Jovovic, Jan 05 2004 a(n) = A034178(4*n - 1). - Michael Somos, May 11 2011 G.f.: Sum_{n >= 1} x^(3*n-2)/(1 - x^(4*n-3)). - Peter Bala, Jan 08 2021 From Amiram Eldar, Dec 26 2022: (Start) a(n) = A000005(A004767(n-1))/2. Sum_{k=1..n} a(k) = (log(n) + 2*gamma - 1 + 4*log(2))*n/4 + O(n^(1/3)*log(n)), where gamma is Euler's constant (A001620). (End) G.f.: Sum_{n >= 1} x^(n^2)/(1-x^(2*n-1)) (conjecture). - Joerg Arndt, Jan 04 2024 EXAMPLE SDT(34) = 4 since we have 34 or 11+12+11 or 6+7+8+7+6 or 2+3+4+5+6+5+4+3+2, Also 4*34 - 1 = 135 = (3^3)*(5^1) so that r1=3 and r2=1 (p1=3 and p2=5), resulting in SDT(34) = (3+1)*(1+1)/2 = 4. a(4) = 2 since 4 = 2^2 - 2*0 = 4^2 - 2*6. Also A034178(4*4 - 1) = 2 since 15 = 4^2 - 1^2 = 8^2 - 7^2. - Michael Somos, May 11 2011 G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + 2*x^7 + x^8 + 2*x^9 + 2*x^10 + x^11 + ... Number of divisors of numbers 3 (mod 4) (see the Jovovic Jan 05 2004 comment): a(16) = 3 from the 2*3 = 6 divisors [1, 3, 7, 9, 21, 63] of 63 = A004767(15), being 1, -1, -1, 1, 1, -1 (mod 4). - Wolfdieter Lang, Jul 29 2016 MATHEMATICA (* This defines SDT(n): *) SDT[n_] := Length[Cases[Range[1, n], j_ /; Cases[Range[1, j], k_ /; Plus @@ Join[Range[k, j], Range[j - 1, k, -1]] == n] != {}]] The restricted factorization technique for obtaining SDT(n) is encoded as follows: SDT[n_] := (Times @@ Cases[FactorInteger[4 n - 1], {p_, r_} -> r + 1])/2 Rest[ CoefficientList[ Series[ Sum[x^k/(1 - x^(4k - 1)), {k, 111}], {x, 0, 110}], x]] (* Robert G. Wilson v, Sep 20 2005 *) a[ n_] := If[ n < 1, 0, With[{m = 4 n - 1}, Sum[1 - Sign@Mod[m - k^2, 2 k], {k, Sqrt@m}]]]; (* Michael Somos, Aug 01 2016 *) a[n_] := DivisorSigma[0, 4*n - 1]/2; Array[a, 100] (* Amiram Eldar, Dec 26 2022 *) PROG (PARI) {a(n) = if( n<1, 0, n = 4*n-1; sum(k=1, sqrtint(n), 0 == (n - k^2) % (2*k)))}; /* Michael Somos, Aug 01 2016 */ CROSSREFS Cf. A000005, A001620, A004767, A001227, A034178, A359227, A359240. Sequence in context: A369179 A319907 A357112 * A090629 A248623 A086412 Adjacent sequences: A078700 A078701 A078702 * A078704 A078705 A078706 KEYWORD nonn,easy AUTHOR R. L. Coffman, K. W. McLaughlin and R. J. Dawson (robert.l.coffman(AT)uwrf.edu), Dec 19 2002 STATUS approved

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