A078714 a(n) = smallest number m which can be obtained in n ways by subtracting twice a triangular number from a perfect square.

1, 4, 16, 34, 142, 79, 1276, 289, 394, 709, 103336, 1024, 930022, 6379, 3544, 2599, 75331762, 5119, 677985856, 9214, 31894, 516679, 54916854316, 12994, 88594, 4650109, 30319, 82924, 40034386796182, 46069, 360309481165636, 33784, 2583394, 376658809, 797344

Offset: 1

Author

R. L. Coffman, K. W. McLaughlin and R. J. Dawson (robert.l.coffman(AT)uwrf.edu), Dec 19 2002

Keywords

• nonn

Comments

The minimum number m (denoted by LSDT(n)) which can be represented in n different ways as a symmetric unimodal consecutive integer sequence (e.g., 6+7+8+7+6) that sums to the integer m. More precisely, n is the number of ways to arrange m objects into symmetrically-placed, congruent isosceles trapezoids adjoined at overlapping largest bases and m is the minimum number of objects that allows this number of arrangements.

a(23)-a(50) are ?, 12994, 88594, 4650109, 30319, 82924, ?, 46069, ?, 33784, 2583394, 376658809, 797344, 78829, ?, ?, 23250544, 148129, ?, 414619, ?, 6716824, 272869, ?, ?, 168919, 19933594, 1151719. - Robert G. Wilson v, Dec 24 2002

Examples

			Let SDT(n) = the number, k, of symmetric double trapezoidal arrangements of n objects, then 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. For SDT(n) = 4, we have n = 34 or 49 or 58 or 64 ..., so that the least value of SDT(n)=4 is LSDT(4) = 34. 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 and 34 is the least value of n which satisfies 4*n-1 so that one half the number of odd divisors equals 4.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..2098 (a(2099) exceeds 1000 digits).
• T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Crossrefs

Cf. A078703, A038547, A018782, A204046, A204086.

Programs

Mathematica

The following function determines the number of ways, SDT(n), of arranging n identical objects into symmetric double trapezoidal arrangements: SDT[n_] := (Times @@ Cases[FactorInteger[4 n - 1], {p_, r_} -> r + 1])/2 The program below computes the first few terms of the sequence LSDT(k)=min{n:SDT(n)=k}. The output is in the form {{1, LSDT(1)}, {2, LSDT(2)}, {3, LSDT(3)}, ...}: Union[Sort[{SDT[ # ], #} & /@ Range[1, 100000]], SameTest -> (#1[[1]] == #2[[1]] &)]

Formula

LSDT(k)={min n: SDT(n)=k}, where SDT(n)=((r1+1)*(r2+1)*...)/2 and ((p1^r1)*(p2^r2)*...) is the factorization of 4n-1 into (odd) primes.

a(n) = (A204086(n) + 1)/4. - Ray Chandler, Jan 10 2012

For odd prime p, a(p) = (3^(p-1)*7 + 1)/4.

Extensions

Missing terms noted in Comments and b-file from Ray Chandler, Jan 10 2012