A394241 Minimal value of Sum_{i=1..n-1} p(i)*p(i+1), as p ranges over all permutations of odd numbers {1,3,...,2*n-1}.

0, 0, 3, 8, 25, 50, 95, 156, 245, 358, 507, 688, 913, 1178, 1495, 1860, 2285, 2766, 3315, 3928, 4617, 5378, 6223, 7148, 8165, 9270, 10475, 11776, 13185, 14698, 16327, 18068, 19933, 21918, 24035, 26280, 28665, 31186, 33855, 36668, 39637, 42758, 46043, 49488

Offset

0,3

Comments

This question originated from generalization of the puzzle 'Low Budget' in the book Mathematical Puzzles and Curiosities.

References

I. David, T. Khovanova, and Y. Shpilman, Mathematical Puzzles and Curiosities, World Scientific, 2026, p. 2.

Links

Table of n, a(n) for n=0..43.

Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Formula

a(n) = (2*n^3 - 6*n^2 + 13*n - 12 + 3*(-1)^n)/3 for n>=2.

a(n) = 4*A026035(n) - 2*n^2 + 3*n - 3.

G.f.: x^2*(2*x^4-3*x^3+7*x^2-x+3)/((x+1)*(x-1)^4).

Example

For n=4, the minimum is achieved for permutation 5,3,1,7 of the first four odd numbers. The value is 5*3+3*1+1*7=25. Thus, a(4) = 25.

Mathematica

a[n_]:= (2*n^3 - 6*n^2 + 13*n - 12 + 3*(-1)^n)/3; a[0]=0; a[1]=0; Array[a, 44, 0] (* or *)
CoefficientList[Series[x^2*(2*x^4-3*x^3+7*x^2-x+3)/((x+1)*(x-1)^4) , {x, 0, 43}], x] (* James C. McMahon, Apr 25 2026 *)

Prog

(Python)
def A394241(n): return n*(n*(n-3<<1)+13)//3-(5 if n&1 else 3) if n>1 else 0 # Chai Wah Wu, Apr 24 2026

Crossrefs

Cf. A026035, A101986, A110610, A110611, A393145, A395196.

Sequence in context: A220486 A180380 A057420 * A076049 A384974 A026970

Adjacent sequences: A394238 A394239 A394240 * A394242 A394243 A394244

Keyword

nonn,easy

Author

Tanya Khovanova and PRIMES STEP junior group, Apr 19 2026

Status

approved