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Partition the j digits of n into blocks of k, with 1 <= k <= j-1, starting at left and multiply. Sum of these numbers equals n.
1

%I #24 Aug 06 2016 23:41:44

%S 396,1064,1518,1545,4318,4345,4563,12154,21154,145250,517150,532315,

%T 43321250,66710504,175401008,407902400,492701500,1148032202,

%U 10144502500,35308402400,44916701500,90751434020,101445025000,353084024000,449167015000,907514340200

%N Partition the j digits of n into blocks of k, with 1 <= k <= j-1, starting at left and multiply. Sum of these numbers equals n.

%C The number of partitions of k digits are [j-(j mod k)]/k. If j is not a multiple of k the last partition has j mod k digits. E.g.: the partitions of 3 digits of 5573670 are 557 and 367 plus a partition of one digit, 0 (here j=7 and k=3). - _Paolo P. Lava_, Aug 04 2016

%e 3*9*6 + 39*6 = 396;

%e 1*0*6*4 + 10*64 + 106*4 = 1064;

%e 1*2*1*5*4 + 12*15*4 + 121*54 + 1215*4 = 12154.

%p P:=proc(q) local a,b,c,d,j,k,n;

%p for n from 1 to q do c:=0; for k from 1 to ilog10(n) do a:=1; b:=n; d:=ilog10(n)+1;

%p for j from 1 to trunc(d/k) do a:=a*(trunc(n/10^(d-j*k)) mod 10^k); od;

%p if d-trunc(d/k)*k>0 then a:=a*(n mod 10^(d-trunc(d/k)*k)); fi;

%p c:=c+a; od; if n=c then print(n); fi; od; end: P(10^6);

%Y Cf. A275170.

%K nonn,base,more

%O 1,1

%A _Paolo P. Lava_, Jul 19 2016

%E a(18)-a(26) from _Giovanni Resta_, Jul 21 2016