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%I #16 Oct 26 2014 04:55:49
%S 1,0,1,0,2,0,1,0,1,0,2,0,1,0,1,0,2,0,3,1,1,0,3,0,1,0,1,0,3,0,1,0,1,1,
%T 1,0,1,0,2,0,2,0,1,0,1,0,2,0,2,0,1,0,3,1,2,0,1,0,3,0,1,0,1,0,2,0,1,1,
%U 1,0,3,0,1,0,1,1,3,0,1,1,1,0,5,0,1,1,1,0,4,0,2,1,1,0,3,0,1,0,3,0
%N Number of bases b for which the base-b alternate digital sum of n is b.
%C For the definition of the alternate digital sum, see A055017 or A225693.
%C For reference: we write altDigitSum_b(x) for the base-b alternate digital sum of x according to A055017.
%C The number of counted bases includes the special base 1. The base-1 expansion of a natural number is defined as 1=1_1, 2=11_1, 3=111_1 and so on. As a result, the base-1 alternate digital sum is 0 if n is even, and is 1 if n is odd.
%C For odd n we have altDigitSum_1(n) = 1, and thus a(n) >= 1.
%C The altDigitSum_b(n) is < b for bases b that satisfy b > b0 := floor((sqrt(4n+5) - 1)/2), and thus a(n) <= b0. This boundary can also be expressed as b0 := floor(sqrt(n - floor(sqrt(n)) + 1)).
%C If n + 1 is an oblong number (see A002378), then b := (sqrt(4n+5) - 1)/2 is an integer equal to b0 and b^2 + b - 1 = n. This implies altDigitSum_b(n) = 1 + 0 + b - 1 = b and shows, that there are infinitely many n with a base b > 1 such that altDigitSum_b(n) = b. It follows a(n) >= 2 for n = 5, 11, 19, 29, 41, ... (since those n are odd).
%C Moreover, a(n) >= 2 is also true for n == b(b+1)-1 (mod (b+1)b^3), b>1.
%C Example 1: altDigitSum_2(n) = 2 for n == 5 (mod 24).
%C Example 2: altDigitSum_3(n) = 3 for n == 11 (mod 108).
%C Example 3: altDigitSum_4(n) = 4 for n == 19 (mod 320).
%C If b is a base such that the base-b alternate digital sum of n is b, then b + 1 is a divisor of n + 1. Thus the number of such bases is also limited by the number of divisors of n + 1 (see formula section).
%C If b + 1 is a divisor of n + 1, then b is not necessarily a base such that base-b alternate digital sum of n is b. Example: 2, 3, 4, 6 and 12 are divisors of 12, and altDigitSum_1(11) = 1, altDigitSum_3(11) = 3, but altDigitSum_2(11) = -1, altDigitSum_5(11) = -1, altDigitSum_11(11) = -1.
%C a(b*n) > (b*n) mod 2 for all b > 1 which satisfy altDigitSum_b(n) = -b.
%C Example 4: altDigitSum_2(10) = -2, hence a(2*10) > 0
%C Example 5: altDigitSum_3(33) = -3, hence a(3*33) > 1
%C The first n with a(n) = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... are n = 1, 5, 19, 89, 83, 359, 419, 1259, 839, 3359, ... .
%H Hieronymus Fischer, <a href="/A239704/b239704.txt">Table of n, a(n) for n = 1..10000</a>
%F a(A002378(n)-1) = a(n^2+n-1) >= 2, for n > 1.
%F a(n) = 0, if n + 1 is a prime.
%F a(n) <= floor(sigma_0(n+1)/2).
%e a(1) = 1, since altDigitSum_1(1) = 1 and altDigitSum_b(1) = 1 < b for all b > 1.
%e a(2) = 0, since altDigitSum_1(2) = 0 (because of 2 = 11_1), and altDigitSum_2(2) = -1 (because of 2 = 10_2), and altDigitSum_b(2) = 2 < b for all b > 2.
%e a(3) = 1, since altDigitSum_1(3) = 1 (because of 3 = 111_1), and altDigitSum_2(3) = 0 (because of 3 = 11_2), and altDigitSum_3(3) = -1 (because of 3 = 10_3), and altDigitSum_b(3) = 3 < b for all b > 3.
%e a(5) = 2, since altDigitSum_1(5) = 1 (because of 5 = 11111_1), and altDigitSum_2(5) = 2 (because of 5 = 101_2), and altDigitSum_3(5) = 1 (because of 5 = 12_3), and altDigitSum_4(5) = 0 (because of 5 = 11_4), and altDigitSum_5(5) = 1 (because of 5 = 10_5), and altDigitSum_b(5) = 5 < b for all b > 5.
%o (Smalltalk)
%o "> Version 1: simple calculation for small numbers.
%o Answer the number of bases b for which the alternate
%o digital sum of n in base b is b. Valid for bases b > 0.
%o Usage: n numOfBasesWithAltDigitalSumEQ0
%o Answer: a(n)"
%o numOfBasesWithAltDigitalSumEQBase
%o | b q numBases |
%o self < 2 ifTrue: [^0].
%o numBases := 1.
%o q := self sqrtTruncated.
%o b := 1.
%o [b < q] whileTrue:[
%o (self altDigitalSumRight: b) = 0
%o ifTrue: [numBases := numBases + 1].
%o b := b + 1].
%o ^numBases
%o [by _Hieronymus Fischer_, May 08 2014]
%o -----------
%o (Smalltalk)
%o "> Version 2: accelerated calculation for large numbers.
%o Answer the number of bases b for which the alternate
%o digital sum of n in base b is b."
%o numOfBasesWithAltDigitalSumEQBase
%o | numBases div b bsize |
%o div := (self + 1) divisors.
%o numBases := 0.
%o bsize := div size // 2 + 1.
%o 2 to: bsize
%o do:
%o [:i |
%o b := (div at: i) - 1.
%o [(self altDigitalSumRight: b) = b
%o ifTrue: [numBases := numBases + 1]]].
%o ^numBases
%o [by _Hieronymus Fischer_, May 08 2014]
%Y Cf. A055017, A225693, A187813.
%Y Cf. A239703, A239705, A239706, A239707.
%Y Cf. A002378, A008864, A000040, A000005.
%K nonn
%O 1,5
%A _Hieronymus Fischer_, May 08 2014
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