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
Keywords
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.
Programs
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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
Comments