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Triangle of octanomial coefficients read by rows: n-th row is obtained by expanding ((1+x)*(1+x^2)*(1+x^4))^n mod 2 and converting to decimal.
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%I #20 Oct 14 2024 02:53:21

%S 1,255,21845,3342387,286331153,64424509455,5519032976645,

%T 844437815230467,72340172838076673,18446744073709551615,

%U 1567973246265311887445,241781474574111093044019,20552052528097949033496593,4660480146812799619066433295,396140812663555408357742346245,61084913312720243968763869790979

%N Triangle of octanomial coefficients read by rows: n-th row is obtained by expanding ((1+x)*(1+x^2)*(1+x^4))^n mod 2 and converting to decimal.

%C A generalization: Denote {a_k(n)}_(n>=0) the sequence of triangle of 2^k-nomial coefficients [read by rows: n-th row is obtained by expanding ((1+x)*(1+x^2)*...*(1+x^(2^(k-1)))^n ] mod 2 converted to decimal. Then a_k(n)=A001317((2^k-1)*n). [Proof is based on the fact (following from the Lucas theorem for the binomial coefficients) that the k-th row of Pascal triangle contains odd coefficients only iff k is Mersenne number (k=2^m-1)].

%F a(n) = A001317(7*n).

%t a = Plus@@(x^Range[0, 7]); Table[FromDigits[Mod[CoefficientList[a^n, x], 2], 2], {n, 0, 15}]

%o (Python)

%o def A177897(n): return sum((bool(~(m:=7*n)&m-k)^1)<<k for k in range(7*n+1)) # _Chai Wah Wu_, May 03 2023

%o (PARI) a(n) = subst(lift(Mod(1+'x, 2)^(7*n)), 'x, 2); \\ _Michel Marcus_, Oct 14 2024

%Y Cf. A001317, A008589, A177882.

%K nonn,easy

%O 0,2

%A _Vladimir Shevelev_, Dec 15 2010

%E More terms from _Michel Marcus_, Oct 14 2024