A257850 a(n) = floor(n/10) * (n mod 10).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0, 8

Offset

0,13

Comments

Equivalently, write n in base 10, multiply the last digit by the number with its last digit removed.

See A142150(n-1) for the base 2 analog and A257843 - A257849 for the base 3 - base 9 variants.

The first 100 terms coincide with those of A035930 (maximal product of any two numbers whose concatenation is n), A171765 (product of digits of n, or 0 for n<10), A257297 ((initial digit of n)*(n with initial digit removed)), but the sequence is of course different from each of these three.

The terms a(10) - a(100) also coincide with those of A007954 (product of decimal digits of n).

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,-1).

Formula

a(n) = 2*a(n-10)-a(n-20). - Colin Barker, May 11 2015

G.f.: x^11*(9*x^8+8*x^7+7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+2*x+1) / ((x-1)^2*(x+1)^2*(x^4-x^3+x^2-x+1)^2*(x^4+x^3+x^2+x+1)^2). - Colin Barker, May 11 2015

Mathematica

Table[Floor[n/10] Mod[n, 10], {n, 100}] (* Vincenzo Librandi, May 11 2015 *)

Prog

(PARI) a(n, b=10)=(n=divrem(n, b))[1]*n[2]
(Magma) [Floor(n/10)*(n mod 10): n in [0..100]]; // Vincenzo Librandi, May 11 2015
(Python) def A257850(n): return n//10*(n%10) # M. F. Hasler, Sep 01 2021

Crossrefs

Cf. A142150 (the base 2 analog), A115273, A257844 - A257849.

Cf. also A007954, A035930, A171765, A257297.

Sequence in context: A004720 A088118 A330633 * A080464 A171765 A257297

Adjacent sequences: A257847 A257848 A257849 * A257851 A257852 A257853

Keyword

nonn,base,easy

Author

M. F. Hasler, May 10 2015

Status

approved