

A326387


Nonoblong composites m such that beta(m) = tau(m)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.


7



15, 21, 26, 40, 57, 62, 80, 85, 86, 91, 93, 111, 114, 124, 129, 133, 146, 170, 171, 172, 183, 215, 219, 222, 228, 242, 259, 266, 285, 292, 312, 314, 333, 341, 343, 365, 366, 381, 399, 422, 438, 444, 455, 468, 471, 482, 507, 518, 532, 549, 553
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OFFSET

1,1


COMMENTS

As tau(m) = 2 * beta(m), the terms of this sequence are not squares.
The number of Brazilian representations of a nonoblong number m with repdigits of length = 2 is beta'(n) = tau(n)/2  1.
This sequence is the first subsequence of A326380: nonoblong composites which have only one Brazilian representation with three digits or more.


LINKS



EXAMPLE

tau(m) = 4 and beta(m) = 2 for m = 15, 21, 26, 57, 62, 85, 86, ... with 15 = 1111_2 = 33_4.
tau(m) = 8 and beta(m) = 4 for m = 40 = 1111_3 = 55_7 = 44_9 = 22_19.
tau(m) = 10 and beta(m) = 5 for m = 80 = 2222_3 = 88_9 = 55_15 = 44_19 = 22_39.


PROG

(PARI) isoblong(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
beta(n) = sum(i=2, n2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(m) = !isprime(m) && !isoblong(m) && (beta(m) == numdiv(m)/2); \\ Michel Marcus, Jul 15 2019


CROSSREFS

Cf. A326386 (nonoblongs with tau(m)/2  1), A326388 (nonoblongs with tau(m)/2 + 1), A326389 (nonoblongs with tau(m)/2 + 2).


KEYWORD

nonn,base


AUTHOR



STATUS

approved



