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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
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
Sequence in context: A369179 A319907 A357112 * A090629 A248623 A086412
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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Last modified May 18 15:59 EDT 2024. Contains 372664 sequences. (Running on oeis4.)