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Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.
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%I #27 Aug 16 2019 12:52:50

%S 16777215,435356467,1073741823,68719476735,1099511627775,

%T 4398046511103,35184372088831,281474976710655,14901161193847656,

%U 18014398509481983

%N Numbers m such that beta(m) = tau(m)/2 + k for some k >= 4, where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

%C As tau(m) = 2 * (beta(m) - k) is even, the terms of this sequence are not squares.

%C There are two classes of terms (see array in link and examples):

%C 1) Non-oblong composites which have five or more Brazilian representations with three digits or more, they form a subsequence of A326705. The smallest example is a(1) = 16777215 = M_24.

%C 2) Oblong numbers that have six or more Brazilian representations with three digits or more, they form a subsequence of A309062. The smallest example is a(9) (see 2nd example).

%C For a(1) to a(10), the numbers k are respectively 5, 4, 5, 6, 5, 5, 4, 7, 4 and 5.

%C Some Mersenne numbers are terms: M_24 = a(1), M_30 = a(3), M_36 = a(4), M_40 = a(5), M_42 = a(6), M_45 = a(7), M_48 = a(8), M_54 = a(10).

%H Bernard Schott, <a href="/A326382/a326382.pdf">Array of relations beta = f(tau)</a>

%H <a href="https://oeis.org/wiki/Index_to_OEIS:_Section_Br#Brazilian_numbers">Index entries for sequences related to Brazilian numbers</a>

%e One example of each type:

%e 1) Non-oblong with beta"(m) = 5; tau(435356467) = 64 and 435356467 = (6^12 - 1)/5 has exactly five Brazilian representations with three digits or more: R(12)_6 = 777777_36 = (43,43,43)_216 = (259,259,259)_1296 = (31,31,31)_3747 and has 31 representations with 2 digits, so beta(435356467) = 36 and k = 4.

%e 2) Oblong with beta"(m) = 6; tau(14901161193847656) = 768 and 14901161193847656 = (5^24 - 1)/4 = 122070312*122070313 is oblong. The six Brazilian representations with three digits or more of this term are R(24)_5 = 666666666666_25 = (31,31,31,31,31,31,31,31)_125 = (156,156,156,156,156)_625, =(3906,3906,3906,3906)_15625 = (97656,97656,97656)_390625 so beta"(14901161193847656) = 6 and beta(61035156) = (tau(61035156)/2 - 2) + 6 = 388 and k = 4.

%o (PARI) okrepu3(b, target, lim) = {my(k = 3, nb = 0, x); while ((x=(b^k-1)/(b-1)) <= target, if (x==target, nb++); k++); nb; }

%o dge3(n, d) = {my(nb=0, ndi, limi); for (i=1, #d, ndi = n/d[i]; limi = sqrtint(ndi); for (k=d[i]+1, limi, nb += okrepu3(k, ndi, limi); ); ); nb; }

%o deq2(n, d) = {my(nb=0, nk); for (k=1, #d\2, nk = (n - d[k])/d[k]; if (nk > d[k], nb++); ); nb; }

%o beta(n) = {if (n<3, return (0)); my(d=divisors(n)); deq2(n, d) + dge3(n, d) - 1; }

%o isok(n) = beta(n) - numdiv(n)/2 > = 4; \\ _Michel Marcus_, Aug 10 2019

%Y Cf. A000005 (tau), A220136 (beta).

%Y Subsequence of A167782, A167783 and A290869.

%Y Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326381 (tau(m)/2 + 1), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3), this sequence (tau(m)/2 + k, k >= 4).

%Y Cf. A291592 (Mersenne numbers).

%K nonn,base,hard,more

%O 1,1

%A _Bernard Schott_, Aug 09 2019