OFFSET
1,1
COMMENTS
Exercise A1 of 33rd Putnam Exam in 1972 asked one to prove that there are no four consecutive binomial coefficients C(m,k), C(m,k+1), C(m,k+2), C(m,k+3) in arithmetic progression (see link and reference).
However, as there exist such progressions with 3 terms, this sequence lists the smallest term of these arithmetic progressions.
Three consecutive binomial coefficients form an arithmetic progression iff m = n^2+4n+2, n >= 1 (2nd comment of A008865), and then, corresponding k = (n^2+3n-2)/2. Successive pairs (m,k) are (7,1), (14,4), (23,8), (34,13), (47,19), ...
By symmetry of Pascal's triangle with C(m,k) = C(m,m-k), there exists another decreasing arithmetic progression with the same 3 terms in the same row.
Corresponding common differences are in A349737.
REFERENCES
G. L. Alexanderson, L. F. Klosinski and L. C. Larson, The William Lowell Putnam Mathematical Competition, Problems and Solutions 1965-1984, The Mathematical Association of America, 1985, page 17.
LINKS
FORMULA
a(n) ~ c*2^(n^2+4*n)/n, where c = 4*sqrt(2/(Pi*e)). - Stefano Spezia, Nov 29 2021
EXAMPLE
For n = 1, row 7 of Pascal's triangle is 1, 7, 21, 35, 35, 21, 7, 1; C(7,1) = 7, C(7,2) = 21 and C(7,3) = 35 form an arithmetic progression with common difference = 14, hence a(3) = 7 = C(7,1).
For n = 2, row 14 is 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1; C(14,4) = 1001 , C(14,5) = 2002 and C(14,6) = 3003 form an arithmetic progression with common difference = 1001, hence a(4) = 1001 = C(14,4).
MAPLE
Sequence = seq(binomial(n^2+4*n+2, (n^2+3*n-2)/2), n=1..16);
MATHEMATICA
nterms=15; Table[Binomial[n^2+4n+2, (n^2+3n-2)/2], {n, nterms}] (* Paolo Xausa, Nov 29 2021 *)
PROG
(PARI) a(n) = binomial(n^2+4*n+2, (n^2+3*n-2)/2) \\ Andrew Howroyd, Oct 29 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bernard Schott, Nov 28 2021
EXTENSIONS
Missing a(9) = 23955...79 inserted by Georg Fischer, Oct 29 2023
STATUS
approved